Q: Q2. A resistance R is connected to a cell of emf E and internal resistance r. The terminal potential difference is V.\n(a) Express V in terms of E, I and r.\n(b) Express V in terms of

(a) V = E - Ir\n(b) V = E * R / (R + r)\n(c) V = IR\n(d) V < E. This is because a potential drop (Ir) occurs across the internal resistance of the cell when current flows through it, reducing the potential difference available across the external resistance.

Q: Q3. (a) The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?\n(b) A batt

(a) The maximum current that can be drawn from the battery is 30 A.\n(b) The resistance of the resistor is 17 Ω. The terminal voltage of the battery when the circuit is closed is 8.5 V.

Q: Q4. (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?\n(b) If the combination is connected to a battery of emf 12 V and neg

(a) The total resistance of the series combination is 6 Ω.\n(b) The potential drop across the 1 Ω resistor is 2 V, across the 2 Ω resistor is 4 V, and across the 3 Ω resistor is 6 V.

Q: Q5. (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in parallel. What is the total resistance of the combination?\n(b) If the combination is connected to a battery of emf 12 V and n

(a) The total resistance of the parallel combination is 6/11 Ω (approximately 0.545 Ω).\n(b) The current through the 1 Ω resistor is 12 A, through the 2 Ω resistor is 6 A, and through the 3 Ω resistor is 4 A. The total current drawn from the battery is 22 A.

Q: Q6. A wire of resistance R is stretched to twice its original length. What will be its new resistance?

When a wire is stretched, its volume remains constant. Let the original length be L, original cross-sectional area be A, and original resistance be R. The new length is L' = 2L. Since the volume remains constant, A'L' = AL, so A' = AL/L' = AL/(2L) = A/2. The new resistance R' can be calculated using the formula R = ρL/A, where ρ is the resistivity of the material.\n\nOriginal resistance: R = ρL/A\nNew resistance: R' = ρL'/A' = ρ(2L)/(A/2) = ρ(4L/A) = 4(ρL/A) = 4R\n\nThus, the new resistance of the wire will be 4R.

Q: Q7. A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

The resistance of a material changes with temperature according to the formula:\nR_T = R_0 [1 + α(T - T_0)]\nwhere R_T is the resistance at temperature T, R_0 is the resistance at a reference temperature T_0, and α is the temperature coefficient of resistivity.\n\nGiven:\nInitial resistance R₁ = 2.1 Ω at T₁ = 27.5 °C\nFinal resistance R₂ = 2.7 Ω at T₂ = 100 °C\n\nWe can use the formula relating two resistances at two different temperatures:\nR₂ = R₁ [1 + α(T₂ - T₁)]\n\nSubstitute the given values:\n2.7 = 2.1 [1 + α(100 - 27.5)]\n2.7 = 2.1 [1 + α(72.5)]\n\nDivide both sides by 2.1:\n2.7 / 2.1 = 1 + 72.5α\n9 / 7 = 1 + 72.5α\n\nSubtract 1 from both sides:\n9/7 - 1 = 72.5α\n(9 - 7)/7 = 72.5α\n2/7 = 72.5α\n\nSolve for α:\nα = (2/7) / 72.5\nα = 2 / (7 × 72.5)\nα = 2 / 507.5\nα ≈ 0.0039408 °C⁻¹\n\nRounding to three significant figures, the temperature coefficient of resistivity of silver is approximately 3.94 × 10⁻³ °C⁻¹.

Q: Q8. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery du

During charging, the current flows into the battery from the external supply. The net EMF in the circuit is the difference between the supply voltage and the battery's EMF. The total resistance includes the series resistor and the battery's internal resistance.\n\n**1. Calculation of Charging Current (I):**\nSupply voltage, V_supply = 120 V\nBattery EMF, E = 8.0 V\nBattery internal resistance, r = 0.5 Ω\nSeries resistor, R_series = 15.5 Ω\n\nTotal resistance in the circuit, R_total = R_series + r = 15.5 Ω + 0.5 Ω = 16.0 Ω\n\nNet EMF driving the current in the circuit = V_supply - E = 120 V - 8.0 V = 112 V\n\nCharging current, I = (Net EMF) / R_total = 112 V / 16.0 Ω = 7.0 A\n\n**2. Calculation of Terminal Voltage (V_terminal) during charging:**\nDuring charging, the terminal voltage of the battery is given by V_terminal = E + Ir\n\nV_terminal = 8.0 V + (7.0 A × 0.5 Ω)\nV_terminal = 8.0 V + 3.5 V\nV_terminal = 11.5 V\n\n**3. Purpose of the series resistor:**\nThe series resistor (15.5 Ω) is crucial in the charging circuit for two main reasons:\n* **Limiting Current:** It limits the charging current to a safe value. Without it, a very large current would flow from the 120 V supply into the 8.0 V battery, potentially damaging the battery due to overheating or overcharging, and also damaging the supply or wires.\n* **Dissipating Excess Voltage:** It drops the excess voltage from the supply (120 V) to a level suitable for charging the battery (which has an EMF of 8.0 V). The potential drop across the series resistor is I * R_series = 7.0 A * 15.5 Ω = 108.5 V. This ensures that the voltage applied across the battery terminals does not exceed its safe charging voltage.

Q: Q10. A 10 m long wire of uniform cross-section and 20 Ω resistance is connected in series with a battery of 3 V and a rheostat. The rheostat is adjusted so that the potential drop acros

**1. Identify Given Values:**\nLength of wire, L = 10 m\nResistance of wire, R_wire = 20 Ω\nBattery EMF, E = 3 V\nPotential drop across the wire, V_wire = 1.0 V\n\n**2. Calculate the Current flowing through the wire:**\nThe current (I) flowing through the wire can be found using Ohm's Law, as we know the potential drop across the wire and its resistance:\nI = V_wire / R_wire\nI = 1.0 V / 20 Ω\nI = 0.05 A\n\n**3. Calculate the Total Resistance of the Circuit:**\nThe wire, the battery, and the rheostat are connected in series. The total resistance of the circuit (R_total) can be found using Ohm's Law for the entire circuit:\nI = E / R_total\nR_total = E / I\nR_total = 3 V / 0.05 A\nR_total = 60 Ω\n\n**4. Calculate the Resistance of the Rheostat:**\nIn a series circuit, the total resistance is the sum of individual resistances:\nR_total = R_wire + R_rheostat\n60 Ω = 20 Ω + R_rheostat\nR_rheostat = 60 Ω - 20 Ω\nR_rheostat = 40 Ω\n\nThe resistance of the rheostat is 40 Ω.

Question 1

Q1. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery du

Accepted Answer

During charging, the terminal voltage of the battery is 11.5 V. The series resistor is used to limit the current drawn from the DC supply, preventing damage to the battery and the supply due to excessive current.

Question 2

Q2. A resistance R is connected to a cell of emf E and internal resistance r. The terminal potential difference is V.
(a) Express V in terms of E, I and r.
(b) Express V in terms of

Accepted Answer

(a) V = E - Ir
(b) V = E * R / (R + r)
(c) V = IR
(d) V < E. This is because a potential drop (Ir) occurs across the internal resistance of the cell when current flows through it, reducing the potential difference available across the external resistance.

Question 3

Q3. (a) The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?
(b) A batt

Accepted Answer

(a) The maximum current that can be drawn from the battery is 30 A.
(b) The resistance of the resistor is 17 Ω. The terminal voltage of the battery when the circuit is closed is 8.5 V.

Question 4

Q4. (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and neg

Accepted Answer

(a) The total resistance of the series combination is 6 Ω.
(b) The potential drop across the 1 Ω resistor is 2 V, across the 2 Ω resistor is 4 V, and across the 3 Ω resistor is 6 V.

Question 5

Q5. (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and n

Accepted Answer

(a) The total resistance of the parallel combination is 6/11 Ω (approximately 0.545 Ω).
(b) The current through the 1 Ω resistor is 12 A, through the 2 Ω resistor is 6 A, and through the 3 Ω resistor is 4 A. The total current drawn from the battery is 22 A.

Question 6

Q6. A wire of resistance R is stretched to twice its original length. What will be its new resistance?

Accepted Answer

When a wire is stretched, its volume remains constant. Let the original length be L, original cross-sectional area be A, and original resistance be R. The new length is L' = 2L. Since the volume remains constant, A'L' = AL, so A' = AL/L' = AL/(2L) = A/2. The new resistance R' can be calculated using the formula R = ρL/A, where ρ is the resistivity of the material.

Original resistance: R = ρL/A
New resistance: R' = ρL'/A' = ρ(2L)/(A/2) = ρ(4L/A) = 4(ρL/A) = 4R

Thus, the new resistance of the wire will be 4R.

Question 7

Q7. A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Accepted Answer

The resistance of a material changes with temperature according to the formula:
R_T = R_0 [1 + α(T - T_0)]
where R_T is the resistance at temperature T, R_0 is the resistance at a reference temperature T_0, and α is the temperature coefficient of resistivity.

Given:
Initial resistance R₁ = 2.1 Ω at T₁ = 27.5 °C
Final resistance R₂ = 2.7 Ω at T₂ = 100 °C

We can use the formula relating two resistances at two different temperatures:
R₂ = R₁ [1 + α(T₂ - T₁)]

Substitute the given values:
2.7 = 2.1 [1 + α(100 - 27.5)]
2.7 = 2.1 [1 + α(72.5)]

Divide both sides by 2.1:
2.7 / 2.1 = 1 + 72.5α
9 / 7 = 1 + 72.5α

Subtract 1 from both sides:
9/7 - 1 = 72.5α
(9 - 7)/7 = 72.5α
2/7 = 72.5α

Solve for α:
α = (2/7) / 72.5
α = 2 / (7 × 72.5)
α = 2 / 507.5
α ≈ 0.0039408 °C⁻¹

Rounding to three significant figures, the temperature coefficient of resistivity of silver is approximately 3.94 × 10⁻³ °C⁻¹.

Question 8

Q8. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery du

Accepted Answer

During charging, the current flows into the battery from the external supply. The net EMF in the circuit is the difference between the supply voltage and the battery's EMF. The total resistance includes the series resistor and the battery's internal resistance.

**1. Calculation of Charging Current (I):**
Supply voltage, V_supply = 120 V
Battery EMF, E = 8.0 V
Battery internal resistance, r = 0.5 Ω
Series resistor, R_series = 15.5 Ω

Total resistance in the circuit, R_total = R_series + r = 15.5 Ω + 0.5 Ω = 16.0 Ω

Net EMF driving the current in the circuit = V_supply - E = 120 V - 8.0 V = 112 V

Charging current, I = (Net EMF) / R_total = 112 V / 16.0 Ω = 7.0 A

**2. Calculation of Terminal Voltage (V_terminal) during charging:**
During charging, the terminal voltage of the battery is given by V_terminal = E + Ir

V_terminal = 8.0 V + (7.0 A × 0.5 Ω)
V_terminal = 8.0 V + 3.5 V
V_terminal = 11.5 V

**3. Purpose of the series resistor:**
The series resistor (15.5 Ω) is crucial in the charging circuit for two main reasons:
*   **Limiting Current:** It limits the charging current to a safe value. Without it, a very large current would flow from the 120 V supply into the 8.0 V battery, potentially damaging the battery due to overheating or overcharging, and also damaging the supply or wires.
*   **Dissipating Excess Voltage:** It drops the excess voltage from the supply (120 V) to a level suitable for charging the battery (which has an EMF of 8.0 V). The potential drop across the series resistor is I * R_series = 7.0 A * 15.5 Ω = 108.5 V. This ensures that the voltage applied across the battery terminals does not exceed its safe charging voltage.

Question 9

Q9. A potentiometer arrangement consists of a cell of emf 1.25 V and internal resistance negligible connected across the wire AB. A second cell of emf 1.0 V and internal resistance 0.5

Accepted Answer

**1. Calculation of Potential Gradient along the wire AB:**
According to the principle of a potentiometer, at the null point, the potential drop across the balancing length of the potentiometer wire is equal to the EMF of the second cell (since it's an open circuit for the second cell at null point).

EMF of the second cell, E₂ = 1.0 V
Balancing length, l = 35.0 cm = 0.35 m

Potential gradient, k = E₂ / l
k = 1.0 V / 0.35 m
k ≈ 2.857 V/m

Rounding to three significant figures, k = 2.86 V/m

**2. Current drawn from the second cell at the null point:**
At the null point, the galvanometer shows zero deflection. This means no current flows through the second cell, the galvanometer, or the series resistance connected with the second cell. Therefore, the current drawn from the second cell at the null point is **0 A**.

**3. Current drawn from the first cell at the null point:**
The first cell (driver cell) provides the current that flows through the potentiometer wire AB to establish the potential gradient. The current drawn from the first cell is independent of the null point condition for the second cell, as long as the primary circuit is closed.

EMF of the first cell, E₁ = 1.25 V
Internal resistance of the first cell is negligible.

Let the resistance of the potentiometer wire AB be R_AB. The question does not provide the resistance of the potentiometer wire AB or its length. However, we can use the potential gradient to find the current if we assume the length of the wire AB is known or if we are asked for the current in the primary circuit. The question states 'A potentiometer arrangement consists of a cell of emf 1.25 V and internal resistance negligible connected across the wire AB'. This implies the first cell drives current through the wire AB.

Let's re-evaluate the question. It asks for the current drawn from the *first cell* at the null point. The current drawn from the first cell is the current flowing through the potentiometer wire AB. To calculate this, we need the total resistance of the primary circuit. The resistance of the wire AB is not given, nor is its length. However, we have the potential gradient.

Potential gradient k = (Current in primary circuit * Resistance per unit length of wire)

Let R_wire be the resistance of the potentiometer wire AB. Let L be the total length of the wire AB. The question states 'A potentiometer arrangement consists of a cell of emf 1.25 V and internal resistance negligible connected across the wire AB'. This implies that the voltage across the wire AB is 1.25 V.

If the total length of the wire AB is L, then the potential gradient k = V_AB / L = E₁ / L.

From part 1, we found k = 2.857 V/m.
So, 2.857 V/m = 1.25 V / L
L = 1.25 V / 2.857 V/m ≈ 0.4375 m = 43.75 cm.

Now, to find the current drawn from the first cell (I₁), we need the resistance of the wire AB (R_AB). The question does not provide R_AB. If we assume the question implies that the entire 1.25V is dropped across the wire AB, and we have the potential gradient, we can't find the current without the total resistance of the wire.

Let's consider the standard setup. The first cell (E₁) is connected across the wire AB. If there are no other resistors in the primary circuit, then the potential difference across AB is E₁.

Potential gradient k = (I₁ * R_AB) / L = I₁ * (R_AB / L)
We know k = 2.857 V/m.
We also know that the potential difference across the entire wire AB is E₁ = 1.25 V.
So, V_AB = I₁ * R_AB = 1.25 V.

If the length of the wire AB is L, then k = V_AB / L = 1.25 V / L.
We found k = 2.857 V/m.
So, L = 1.25 V / 2.857 V/m = 0.4375 m.

Now, we need R_AB to find I₁.
If the question implies that the potential gradient is established by the first cell across the entire length of the wire, and the potential drop across the entire wire is 1.25 V, then the current in the primary circuit (I₁) is I₁ = E₁ / R_AB. Without R_AB, we cannot calculate I₁.

However, if the question implies that the potential gradient is established by the first cell across the wire AB, and the *total length* of the potentiometer wire is not given, we can only calculate the current if we assume the resistance per unit length of the wire. Since it's not given, there might be an implicit assumption or missing information.

Let's re-read carefully: 'A potentiometer arrangement consists of a cell of emf 1.25 V and internal resistance negligible connected across the wire AB.' This means the potential difference across the entire wire AB is 1.25 V. The potential gradient k = 2.857 V/m. If the total length of the wire is L, then k = 1.25 V / L. This gives L = 1.25 / 2.857 = 0.4375 m.

To find the current drawn from the first cell, we need the resistance of the wire AB. The resistance of the wire AB is not given. This part of the question cannot be answered without the resistance of the potentiometer wire AB.

**Correction/Assumption:** Often in such problems, if the resistance of the potentiometer wire is not given, it's assumed that the question is only asking for the potential gradient and the current from the second cell. If the current from the first cell is asked, it usually implies that the resistance of the wire or resistance per unit length is provided, or the total resistance of the primary circuit is given. Since it's not, I will state that it cannot be calculated without the resistance of the wire AB.

**However, let's consider a common interpretation for such problems:** If the question intends for a calculable answer for the current from the first cell, it might be implicitly asking for the current *through the potentiometer wire* given the potential gradient and the voltage across the wire. But this still requires the total resistance of the wire.

Let's assume there's a typo or missing information for the resistance of the wire AB. If we *hypothetically* assume the total length of the wire is 1 meter (a common length for potentiometer wires), then the resistance of the wire would be R_AB = V_AB / I₁. But we don't have I₁.

Let's stick to the direct interpretation: we have k = 2.857 V/m. The potential difference across the entire wire AB is 1.25 V. The current drawn from the first cell is the current flowing through the wire AB. Let R_AB be the resistance of the wire AB. Then I₁ = E₁ / R_AB = 1.25 V / R_AB. Since R_AB is unknown, I₁ cannot be determined numerically.

**Final Answer for Current from first cell:** Cannot be determined without the resistance of the potentiometer wire AB.

Question 10

Q10. A 10 m long wire of uniform cross-section and 20 Ω resistance is connected in series with a battery of 3 V and a rheostat. The rheostat is adjusted so that the potential drop acros

Accepted Answer

**1. Identify Given Values:**
Length of wire, L = 10 m
Resistance of wire, R_wire = 20 Ω
Battery EMF, E = 3 V
Potential drop across the wire, V_wire = 1.0 V

**2. Calculate the Current flowing through the wire:**
The current (I) flowing through the wire can be found using Ohm's Law, as we know the potential drop across the wire and its resistance:
I = V_wire / R_wire
I = 1.0 V / 20 Ω
I = 0.05 A

**3. Calculate the Total Resistance of the Circuit:**
The wire, the battery, and the rheostat are connected in series. The total resistance of the circuit (R_total) can be found using Ohm's Law for the entire circuit:
I = E / R_total
R_total = E / I
R_total = 3 V / 0.05 A
R_total = 60 Ω

**4. Calculate the Resistance of the Rheostat:**
In a series circuit, the total resistance is the sum of individual resistances:
R_total = R_wire + R_rheostat
60 Ω = 20 Ω + R_rheostat
R_rheostat = 60 Ω - 20 Ω
R_rheostat = 40 Ω

The resistance of the rheostat is 40 Ω.

Question 11

Q11. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery du

Accepted Answer

During charging, the current flows into the positive terminal of the battery. The terminal voltage of the battery is given by V = E + Ir. The purpose of the series resistor is to limit the charging current to a safe value, preventing damage to the battery and the DC supply.

**Calculations:**
Given:
EMF of battery, E = 8.0 V
Internal resistance of battery, r = 0.5 Ω
DC supply voltage, V_supply = 120 V
Series resistor, R_series = 15.5 Ω

1.  **Calculate the net voltage in the circuit:**
    The DC supply voltage opposes the battery's EMF during charging.
    Net voltage = V_supply - E = 120 V - 8.0 V = 112 V

2.  **Calculate the total resistance in the circuit:**
    Total resistance = R_series + r = 15.5 Ω + 0.5 Ω = 16.0 Ω

3.  **Calculate the charging current (I):**
    Using Ohm's Law, I = Net voltage / Total resistance
    I = 112 V / 16.0 Ω = 7.0 A

4.  **Calculate the terminal voltage of the battery during charging (V):**
    During charging, V = E + Ir
    V = 8.0 V + (7.0 A × 0.5 Ω)
    V = 8.0 V + 3.5 V
    V = 11.5 V

**Purpose of the series resistor:**
The series resistor limits the charging current. Without it, the large potential difference (V_supply - E) across the very small total resistance (r) would lead to an excessively high charging current (I = (120 - 8) / 0.5 = 112 / 0.5 = 224 A). Such a high current could:
*   Damage the battery due to overheating or overcharging.
*   Damage the DC supply.
*   Cause a fire hazard.
Therefore, the series resistor ensures that the charging current remains within safe limits for the battery and the power supply.

Question 12

Q12. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63

Accepted Answer

The principle of a potentiometer states that at the balance point, the potential drop across the balanced length of the potentiometer wire is equal to the EMF of the cell being measured. If 'k' is the potential gradient along the potentiometer wire (potential drop per unit length), then EMF (E) is directly proportional to the balancing length (l).

E = k * l

**Calculations:**
Given:
EMF of the first cell, E₁ = 1.25 V
Balancing length for the first cell, l₁ = 35.0 cm
Balancing length for the second cell, l₂ = 63.0 cm

Let E₂ be the EMF of the second cell.

From the potentiometer principle:
E₁ = k * l₁  --- (1)
E₂ = k * l₂  --- (2)

Dividing equation (2) by equation (1):
E₂ / E₁ = (k * l₂) / (k * l₁)
E₂ / E₁ = l₂ / l₁

Rearranging to find E₂:
E₂ = E₁ * (l₂ / l₁)

Substitute the given values:
E₂ = 1.25 V * (63.0 cm / 35.0 cm)
E₂ = 1.25 V * (63 / 35)
E₂ = 1.25 V * 1.8
E₂ = 2.25 V

The emf of the second cell is 2.25 V.

Question 13

Q13. A resistance of R Ω draws a current from a potentiometer. The potentiometer has a total resistance R₀ Ω (Fig. 3.25). A voltage V is supplied to the potentiometer. Derive an express

Accepted Answer

Let the total resistance of the potentiometer wire be R₀. A voltage V is supplied across the ends of the potentiometer wire. The sliding contact is in the middle, meaning the potentiometer wire is divided into two equal halves, each with resistance R₀/2.

**Circuit Description:**
*   The potentiometer wire has total resistance R₀. Let its ends be A and B. A voltage V is applied across A and B.
*   The sliding contact C is at the middle of the wire. So, the resistance of the segment AC is R_AC = R₀/2, and the resistance of the segment CB is R_CB = R₀/2.
*   The external resistance R is connected between point A and the sliding contact C. This means R is in parallel with the segment AC of the potentiometer wire.

**Derivation:**
1.  **Resistance of the parallel combination (R_p):**
    The external resistance R is in parallel with the segment AC, which has resistance R₀/2.
    1/R_p = 1/R + 1/(R₀/2)
    1/R_p = 1/R + 2/R₀
    1/R_p = (R₀ + 2R) / (R * R₀)
    R_p = (R * R₀) / (R₀ + 2R)

2.  **Total resistance of the circuit (R_total):**
    The parallel combination (R_p) is in series with the remaining part of the potentiometer wire, segment CB, which has resistance R₀/2.
    R_total = R_p + R_CB
    R_total = (R * R₀) / (R₀ + 2R) + R₀/2
    R_total = [2 * (R * R₀) + R₀ * (R₀ + 2R)] / [2 * (R₀ + 2R)]
    R_total = [2RR₀ + R₀² + 2RR₀] / [2(R₀ + 2R)]
    R_total = [R₀² + 4RR₀] / [2(R₀ + 2R)]
    R_total = R₀(R₀ + 4R) / [2(R₀ + 2R)]

3.  **Total current drawn from the supply (I_total):**
    I_total = V / R_total
    I_total = V / [R₀(R₀ + 4R) / (2(R₀ + 2R))]
    I_total = [2V(R₀ + 2R)] / [R₀(R₀ + 4R)]

4.  **Voltage across R (V_R):**
    The voltage across R is the voltage across the parallel combination (R_p).
    V_R = I_total * R_p
    V_R = ([2V(R₀ + 2R)] / [R₀(R₀ + 4R)]) * [(R * R₀) / (R₀ + 2R)]
    V_R = [2V * R * R₀ * (R₀ + 2R)] / [R₀ * (R₀ + 4R) * (R₀ + 2R)]
    V_R = (2V * R) / (R₀ + 4R)

Thus, the expression for the voltage across R when the sliding contact is in the middle of the potentiometer is **V_R = (2VR) / (R₀ + 4R)**.

Question 14

Q14. A resistance R is connected to a cell of emf E and internal resistance r. The potential difference across the resistance R is V. Plot a graph between V and R.

Accepted Answer

When a resistance R is connected to a cell of emf E and internal resistance r, the current (I) flowing through the circuit is given by Ohm's law for the entire circuit:

I = E / (R + r)

The potential difference (V) across the external resistance R is given by:

V = I * R

Substituting the expression for I:

V = [E / (R + r)] * R

V = E * R / (R + r)

To plot the graph between V and R, we analyze the behavior of V for different values of R:

1.  **When R = 0 (short circuit):**
    V = E * 0 / (0 + r) = 0 V
    This means when the external resistance is zero, the terminal potential difference across it is also zero.

2.  **When R approaches infinity (open circuit):**
    As R → ∞, we can divide the numerator and denominator by R:
    V = E / (1 + r/R)
    As R → ∞, r/R → 0.
    So, V → E / (1 + 0) = E
    This means when the external resistance is very large (approaching open circuit), the terminal potential difference across it approaches the emf of the cell.

3.  **Intermediate values of R:**
    As R increases from 0, V also increases from 0. The rate of increase of V gradually decreases as R becomes larger. The function V(R) = ER/(R+r) is a monotonically increasing function that saturates at E.

**Graph Description:**
The graph of V versus R will be a curve starting from the origin (0,0) and asymptotically approaching a horizontal line at V = E as R tends to infinity. The curve shows that the terminal potential difference across the external resistance increases with increasing external resistance, eventually reaching the emf of the cell when the external resistance is much larger than the internal resistance.

**Axes:**
*   X-axis: External Resistance (R) in Ohms (Ω)
*   Y-axis: Terminal Potential Difference (V) in Volts (V)

**Shape:** A non-linear, increasing curve, concave down, starting from the origin and asymptotically approaching E.

Question 15

Q15. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC source using a series resistor of 15.5 Ω. What is the terminal voltage of the battery du

Accepted Answer

### Part 1: Terminal voltage of the battery during charging

**Given:**
*   EMF of the battery, E_battery = 8.0 V
*   Internal resistance of the battery, r_battery = 0.5 Ω
*   DC source voltage, V_source = 120 V
*   Series resistor, R_series = 15.5 Ω

**Circuit Description:**
During charging, the DC source (120 V) drives current through the series resistor and the battery. The battery's emf opposes the charging current. The total effective voltage driving the current in the circuit is the difference between the source voltage and the battery's emf.

**1. Calculate the total resistance in the circuit:**
Total resistance, R_total = R_series + r_battery
R_total = 15.5 Ω + 0.5 Ω = 16.0 Ω

**2. Calculate the net voltage driving the current:**
Net voltage, V_net = V_source - E_battery
V_net = 120 V - 8.0 V = 112 V

**3. Calculate the charging current (I):**
Using Ohm's Law, I = V_net / R_total
I = 112 V / 16.0 Ω = 7.0 A

**4. Calculate the terminal voltage of the battery during charging (V_terminal):**
During charging, the terminal voltage across the battery is the sum of its emf and the voltage drop across its internal resistance (since current flows into the positive terminal).
V_terminal = E_battery + I * r_battery
V_terminal = 8.0 V + (7.0 A * 0.5 Ω)
V_terminal = 8.0 V + 3.5 V
V_terminal = 11.5 V

Alternatively, the terminal voltage can also be calculated as the voltage across the battery when the source voltage drops across the series resistor and the battery itself:
V_terminal = V_source - I * R_series
V_terminal = 120 V - (7.0 A * 15.5 Ω)
V_terminal = 120 V - 108.5 V
V_terminal = 11.5 V

Therefore, the terminal voltage of the battery during charging is **11.5 V**.

### Part 2: Purpose of the series resistor

The purpose of having a series resistor (15.5 Ω) in the charging circuit is to **limit the charging current**.

Without the series resistor, the total resistance in the circuit would only be the internal resistance of the battery (0.5 Ω). The charging current would then be:
I_without_resistor = (V_source - E_battery) / r_battery
I_without_resistor = (120 V - 8.0 V) / 0.5 Ω = 112 V / 0.5 Ω = 224 A

Such a high current (224 A) would be extremely dangerous and damaging to the battery and the charging circuit. It could lead to:
*   **Overheating:** Excessive heat generation in the battery and wires.
*   **Damage to the battery:** Permanent damage to the battery plates, electrolyte boiling, or even explosion.
*   **Damage to the DC source:** Overload and potential failure of the charging source.

The series resistor effectively increases the total resistance of the circuit, thereby reducing the charging current to a safe and manageable level (7.0 A in this case). This protects the battery from overcurrent and ensures a controlled charging process.

Question 16

Q16. A resistance R is connected to a cell of emf E and internal resistance r. The potential difference across the resistance R is V.
(a) Write the relation between E, V, R and r.
(b)

Accepted Answer

(a) When a resistance R is connected to a cell of emf E and internal resistance r, a current I flows through the circuit. The total resistance of the circuit is R_total = R + r. According to Ohm's law, the current flowing is I = E / (R + r).

The potential difference across the external resistance R is V = I * R.
Substituting the expression for I:
V = [E / (R + r)] * R
V = E * R / (R + r)

Alternatively, the terminal potential difference V is also given by V = E - I * r. Since I = V/R, we can substitute this into the equation:
V = E - (V/R) * r
V * R = E * R - V * r
V * R + V * r = E * R
V * (R + r) = E * R
V = E * R / (R + r)
This is the relation between E, V, R, and r.

(b) The terminal potential difference V is equal to E (the emf of the cell) when no current is drawn from the cell. This happens when the external circuit is open (i.e., R is infinite), or when the cell is not connected to any external load. In this case, the current I = 0, and from the relation V = E - Ir, we get V = E.

(c) The terminal potential difference V is less than E when current is drawn from the cell (i.e., the cell is discharging). In this case, I > 0, and since V = E - Ir, and Ir is a positive voltage drop across the internal resistance, V will be less than E. This is the most common scenario when a cell is used to power a device.

(d) The terminal potential difference V is greater than E when the cell is being charged. In this scenario, an external source of emf (greater than E) forces current into the cell against its emf. The current flows in the opposite direction compared to discharge. If the charging current is I_c, then the terminal potential difference across the cell during charging is V = E + I_c * r. Since I_c * r is a positive term, V will be greater than E.

Question 17

Q17. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC source using a series resistor of 15.5 Ω. What is the terminal voltage of the battery du

Accepted Answer

During charging, the current flows into the battery. The net EMF in the circuit is the difference between the charging source voltage and the battery's EMF. The total resistance is the sum of the series resistor and the battery's internal resistance.

**1. Calculation of Charging Current (I):**
The charging circuit consists of the 120 V DC source, the series resistor (R_s), and the battery (with its emf ε and internal resistance r). The source voltage drives the current against the battery's emf.
Net EMF in the circuit = V_source - ε_battery = 120 V - 8.0 V = 112 V
Total resistance in the circuit = R_s + r = 15.5 Ω + 0.5 Ω = 16.0 Ω
Using Ohm's Law, I = Net EMF / Total Resistance = 112 V / 16.0 Ω = 7.0 A

**2. Calculation of Terminal Voltage of the Battery (V_terminal) during charging:**
During charging, the terminal voltage of the battery is given by V_terminal = ε_battery + Ir
Where ε_battery is the emf of the battery, I is the charging current, and r is the internal resistance of the battery.
V_terminal = 8.0 V + (7.0 A × 0.5 Ω) = 8.0 V + 3.5 V = 11.5 V

**3. Purpose of the Series Resistor:**
The series resistor (15.5 Ω) is crucial for controlling the charging current. Without it, the total resistance in the circuit would only be the internal resistance of the battery (0.5 Ω). In that case, the charging current would be I = 112 V / 0.5 Ω = 224 A. Such a high current would:
*   Damage the battery due to excessive heat generation.
*   Damage the charging source.
*   Cause rapid gassing and potential explosion of the battery.
*   Lead to inefficient charging and reduced battery life.
Thus, the series resistor limits the charging current to a safe and manageable level (7.0 A in this case), preventing damage to the battery and the charging source.

Question 18

Q18. A resistance R is connected to a cell of emf ε and internal resistance r. The terminal voltage of the cell is V.
(a) Obtain the expression for R in terms of ε, V and r.
(b) For a

Accepted Answer

**(a) Expression for R in terms of ε, V and r:**
When a resistance R is connected to a cell of EMF ε and internal resistance r, a current I flows through the circuit. The terminal voltage V across the external resistance R is given by Ohm's law:

1.  **V = IR** (Equation 1)

The terminal voltage V is also related to the cell's EMF and internal resistance by:

2.  **V = ε - Ir** (Equation 2)

From Equation 1, we can express the current I as:

3.  **I = V/R** (Equation 3)

Substitute Equation 3 into Equation 2:

4.  **V = ε - (V/R)r**

Now, we need to solve for R:

5.  **V + (V/R)r = ε**

6.  **V(1 + r/R) = ε**

7.  **V((R + r)/R) = ε**

8.  **R + r = εR/V**

9.  **R - εR/V = -r**

10. **R(1 - ε/V) = -r**

11. **R((V - ε)/V) = -r**

12. **R = -rV / (V - ε)**

13. **R = rV / (ε - V)**

This is the expression for R in terms of ε, V, and r.

**(b) Dependence of Terminal Voltage V on Current I and V-I Graph:**
The terminal voltage V of a cell is related to the current I drawn from it by the equation:

**V = ε - Ir**

This equation is in the form of a straight line, y = mx + c, where:
*   y = V (terminal voltage)
*   x = I (current)
*   m = -r (slope, which is negative and equal to the negative of the internal resistance)
*   c = ε (y-intercept, which is the EMF of the cell)

**Dependence:**
*   When no current is drawn (I = 0), the terminal voltage V is equal to the EMF (V = ε). This is the open-circuit voltage.
*   As the current I drawn from the cell increases, the term 'Ir' (voltage drop across internal resistance) increases. Since this term is subtracted from the EMF, the terminal voltage V decreases linearly with increasing current I.
*   The rate of decrease of V with respect to I is constant and equal to the internal resistance 'r' of the cell (magnitude of the slope).

**Graph between V and I:**
The graph of V versus I is a straight line with a negative slope.
*   The V-intercept (where I = 0) is ε (the EMF of the cell).
*   The I-intercept (where V = 0) occurs when ε - Ir = 0, which means I = ε/r. This current is the maximum current that can be drawn from the cell (short-circuit current).
*   The slope of the line is -r.

**(c) Maximum current that can be drawn from the cell:**
The maximum current that can be drawn from the cell occurs when the external resistance R is zero. This situation is known as a short circuit.

In a short circuit, the terminal voltage V across the external resistance becomes zero (since R=0, V=IR=0).

Using the equation V = ε - Ir, and setting V = 0:

0 = ε - I_max * r

I_max * r = ε

**I_max = ε / r**

So, the maximum current that can be drawn from the cell is equal to the EMF of the cell divided by its internal resistance.

Question 19

Q19. A 0.5 A current is drawn by a primary cell for 4 minutes. The terminal voltage of the cell is 1.2 V.
(a) Calculate the work done by the cell.
(b) Calculate the total energy spent

Accepted Answer

Given:
Current (I) = 0.5 A
Time (t) = 4 minutes = 4 × 60 seconds = 240 s
Terminal voltage (V) = 1.2 V
EMF (ε) = 1.5 V

**(a) Calculate the work done by the cell (on the external circuit):**
The work done by the cell on the external circuit is the energy delivered to the external resistance. This is calculated using the terminal voltage.
Work done (W) = V × I × t
W = 1.2 V × 0.5 A × 240 s
W = 0.6 × 240 J
W = 144 J

**(b) Calculate the total energy spent by the cell:**
The total energy spent by the cell is the total energy produced by the cell, which includes the energy delivered to the external circuit and the energy dissipated internally due to its internal resistance. This is calculated using the EMF of the cell.
Total energy spent (E_total) = ε × I × t
E_total = 1.5 V × 0.5 A × 240 s
E_total = 0.75 × 240 J
E_total = 180 J

**(c) Calculate the internal resistance of the cell if its emf is 1.5 V:**
The relationship between terminal voltage (V), EMF (ε), current (I), and internal resistance (r) is given by:
V = ε - Ir
We need to find 'r'. Rearranging the formula:
Ir = ε - V
r = (ε - V) / I
r = (1.5 V - 1.2 V) / 0.5 A
r = 0.3 V / 0.5 A
r = 0.6 Ω

Question 20

Q20. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V DC supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery du

Accepted Answer

During charging, the current flows into the positive terminal of the battery. The terminal voltage of the battery is the sum of its EMF and the voltage drop across its internal resistance. The series resistor limits the charging current to a safe value, preventing damage to the battery and the DC supply.

**Calculations:**
1.  **Total EMF in the circuit:** The DC supply (120 V) drives the current against the battery's EMF (8.0 V).
    Net EMF, E_net = V_supply - E_battery = 120 V - 8.0 V = 112 V

2.  **Total resistance in the circuit:** This includes the battery's internal resistance and the series resistor.
    R_total = r_battery + R_series = 0.5 Ω + 15.5 Ω = 16.0 Ω

3.  **Charging current (I):** Using Ohm's Law, I = E_net / R_total
    I = 112 V / 16.0 Ω = 7.0 A

4.  **Terminal voltage of the battery (V_terminal) during charging:** During charging, the terminal voltage is given by V_terminal = E_battery + I * r_battery
    V_terminal = 8.0 V + (7.0 A * 0.5 Ω) = 8.0 V + 3.5 V = 11.5 V

**Purpose of the series resistor:**
The series resistor is crucial in the charging circuit for the following reasons:
*   **Current Limiting:** It limits the charging current to a safe value. Without it, a very large current would flow from the 120 V supply into the 8.0 V battery (due to the large potential difference and small internal resistance), potentially damaging the battery (overheating, electrolyte boiling) and the DC supply.
*   **Voltage Drop:** It causes a significant voltage drop, ensuring that the voltage applied across the battery terminals does not exceed a safe charging voltage.

Question 21

Q21. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63

Accepted Answer

The principle of a potentiometer states that at the balance point, the EMF of the cell is directly proportional to the balancing length of the potentiometer wire. Mathematically, E ∝ L, or E = kL, where k is the potential gradient of the potentiometer wire.

Given:
*   EMF of the first cell (E1) = 1.25 V
*   Balancing length for the first cell (L1) = 35.0 cm
*   Balancing length for the second cell (L2) = 63.0 cm

We need to find the EMF of the second cell (E2).

From the principle of potentiometer:
E1 / E2 = L1 / L2

Rearranging to find E2:
E2 = E1 * (L2 / L1)

Substituting the given values:
E2 = 1.25 V * (63.0 cm / 35.0 cm)
E2 = 1.25 V * 1.8
E2 = 2.25 V

Therefore, the EMF of the second cell is 2.25 V.

Question 22

Q22. The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10^28 m^-3. How long does an electron take to drift from one end of a wire 3.0 m long t

Accepted Answer

To find the time taken for an electron to drift from one end of the wire to the other, we first need to calculate the drift velocity of the electrons. The relationship between current (I), number density of free electrons (n), charge of an electron (e), cross-sectional area (A), and drift velocity (v_d) is given by:
I = n e A v_d

Given values:
*   Number density of free electrons (n) = 8.5 × 10^28 m^-3
*   Length of the wire (L) = 3.0 m
*   Cross-sectional area (A) = 2.0 × 10^-6 m^2
*   Current (I) = 3.0 A
*   Charge of an electron (e) = 1.6 × 10^-19 C

**Step 1: Calculate the drift velocity (v_d)**
From the formula I = n e A v_d, we can express v_d as:
v_d = I / (n e A)

Substitute the given values:
v_d = 3.0 A / ( (8.5 × 10^28 m^-3) × (1.6 × 10^-19 C) × (2.0 × 10^-6 m^2) )
v_d = 3.0 / (8.5 × 1.6 × 2.0 × 10^(28 - 19 - 6))
v_d = 3.0 / (27.2 × 10^3)
v_d = 3.0 / 27200
v_d ≈ 1.103 × 10^-4 m/s

**Step 2: Calculate the time taken (t)**
The time taken for an electron to drift a length L with drift velocity v_d is given by:
t = L / v_d

Substitute the values:
t = 3.0 m / (1.103 × 10^-4 m/s)
t ≈ 2.72 × 10^4 s

To convert this time into hours:
t = 2.72 × 10^4 s / (3600 s/hour)
t ≈ 7.56 hours

Therefore, an electron takes approximately 2.72 × 10^4 seconds or about 7.56 hours to drift from one end of the 3.0 m long wire to its other end.

Question 23

Q23. The earth’s surface has a negative surface charge density of 10^-9 C m^-2. The potential difference between the top of the atmosphere (at about 60 km height) and the surface is 400

Accepted Answer

**Part 1: Calculate the total current if the atmosphere conducts**

Given:
*   Potential difference (V) = 400 kV = 400 × 10^3 V
*   Height of atmosphere (d) = 60 km = 60 × 10^3 m
*   Conductivity of atmosphere (σ) = 10^-14 S m^-1
*   Radius of Earth (R_earth) = 6.4 × 10^6 m

1.  **Calculate the average electric field (E) in the atmosphere:**
    Assuming a uniform electric field between the surface and the top of the atmosphere:
    E = V / d
    E = (400 × 10^3 V) / (60 × 10^3 m)
    E = 400 / 60 V/m = 20 / 3 V/m ≈ 6.67 V/m

2.  **Calculate the current density (J) in the atmosphere:**
    Using Ohm's law in its microscopic form: J = σE
    J = (10^-14 S m^-1) × (20/3 V/m)
    J = (20/3) × 10^-14 A/m^2 ≈ 6.67 × 10^-15 A/m^2

3.  **Calculate the surface area of the Earth (A):**
    The current flows perpendicular to the Earth's surface. So, the area through which the current flows is the surface area of the Earth.
    A = 4πR_earth^2
    A = 4 × 3.14159 × (6.4 × 10^6 m)^2
    A = 4 × 3.14159 × 40.96 × 10^12 m^2
    A ≈ 5.147 × 10^14 m^2

4.  **Calculate the total current (I):**
    I = J × A
    I = (6.67 × 10^-15 A/m^2) × (5.147 × 10^14 m^2)
    I ≈ 3.43 A

So, the total current due to the conduction of the atmosphere is approximately **3.43 A**.

**Part 2: Is the earth’s electric field maintained by the thunderstorm alone?**

Given:
*   Surface charge density (σ_charge) = 10^-9 C m^-2 (negative)
*   Time duration of storm (t_storm) = 0.5 hour = 0.5 × 3600 s = 1800 s

1.  **Calculate the total charge on the Earth's surface (Q_earth):**
    Q_earth = σ_charge × A
    Q_earth = (10^-9 C m^-2) × (5.147 × 10^14 m^2)
    Q_earth = 5.147 × 10^5 C
    (The negative sign indicates the charge is negative, but for magnitude comparison, we use the absolute value).

2.  **Calculate the rate at which charge is being neutralized by atmospheric conduction:**
    The current calculated in Part 1 (I ≈ 3.43 A) represents the rate at which positive charge flows from the atmosphere to the Earth, or negative charge flows from the Earth to the atmosphere, effectively neutralizing the Earth's negative charge.

3.  **Estimate the charge supplied by a typical thunderstorm:**
    A typical thunderstorm can transfer charge of the order of 10 C to 100 C in a short burst. If a storm lasts for half an hour, it might involve multiple lightning strikes. Let's assume a continuous current from a thunderstorm. The question implies whether the *entire* global electric field is maintained by *thunderstorms alone*.

4.  **Compare the charge neutralization rate with potential charge replenishment by thunderstorms:**
    The total current of 3.43 A means that 3.43 Coulombs of charge are neutralized every second. Over the entire Earth, this current continuously tries to neutralize the Earth's negative charge.
    If the Earth's electric field were to be maintained, there must be a continuous mechanism to replenish this charge. Thunderstorms are known to transfer negative charge to the Earth's surface (via lightning) and positive charge to the upper atmosphere, thus acting as a generator that maintains the potential difference. The average current from all thunderstorms globally is estimated to be around 1800 A.

The calculated atmospheric conduction current (3.43 A) is the leakage current that tends to neutralize the Earth's charge. To maintain the Earth's electric field, this leakage current must be continuously replenished. Thunderstorms, through their charge separation and lightning activity, are indeed the primary mechanism for this replenishment. However, a single storm lasting half an hour (which might transfer a few tens to hundreds of Coulombs) cannot maintain the global electric field against a continuous leakage current of 3.43 A across the entire Earth's surface. It's the *collective* action of all global thunderstorms (estimated to be around 1800 A total current) that maintains the Earth's electric field, not a single thunderstorm alone.

**Conclusion:** No, the Earth's electric field is not maintained by a single thunderstorm alone. It is maintained by the collective action of all thunderstorms occurring globally, which continuously replenish the charge lost due to atmospheric conduction.

Question 24

Q24. A resistance of 𝑅 Ω draws a current from a potentiometer. The potentiometer has a total resistance 𝑅₀ Ω. A voltage 𝑉 is supplied to the potentiometer. Derive an expression for t

Accepted Answer

To derive the expression for the voltage across resistance R when the sliding contact is in the middle of the potentiometer, we need to analyze the circuit formed by the potentiometer and the external resistance R. The potentiometer acts as a voltage divider, and when an external load is connected, it modifies the effective resistance of the circuit.

Question 25

Q25. A 10 m long wire of uniform cross-section and 20 Ω resistance is connected in series with a battery of 3 V and a series resistance of 10 Ω. If a galvanometer is connected across 60

Accepted Answer

To find the current through the galvanometer, we first need to determine the resistance of the 60 cm segment of the potentiometer wire. Then, we analyze the circuit with the galvanometer shunting this segment, calculate the equivalent resistance, and finally apply current divider rule or Ohm's law.

Question 26

Q26. A potentiometer wire of length 1 m has a resistance of 10 Ω. It is connected to a 6 V battery in series with a resistance of 5 Ω. Determine the emf of the primary cell which gives

Accepted Answer

To determine the EMF of the primary cell, we need to first calculate the potential gradient of the potentiometer wire. The potential gradient is the potential drop per unit length along the wire. Once the potential gradient is known, the EMF of the cell can be found by multiplying the potential gradient by the balance length.

Chapter 3: Current Electricity