Question 1

Q1. A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?

Accepted Answer

For a purely resistive AC circuit, we can apply Ohm's law using RMS values to find the current. The net power consumed is calculated using the formula P = V_rms * I_rms, as the power factor for a resistor is 1.

Given:
Resistance, R = 100 Ω
RMS voltage, V_rms = 220 V
Frequency, f = 50 Hz

(a) The rms value of current (I_rms) is calculated using Ohm's law:
I_rms = V_rms / R
I_rms = 220 V / 100 Ω = 2.2 A

(b) The net power consumed over a full cycle is given by:
P = V_rms * I_rms
P = 220 V * 2.2 A = 484 W
Alternatively, P = (I_rms)² * R = (2.2 A)² * 100 Ω = 4.84 * 100 = 484 W.

Question 2

Q2. (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?

Accepted Answer

The relationship between the peak value (maximum value) and the root mean square (rms) value for a sinusoidal alternating current or voltage is a fundamental concept.

(a) To find the rms voltage (V_rms) from the peak voltage (V_m):
The formula is V_rms = V_m / √2.
Given V_m = 300 V.
V_rms = 300 / √2 ≈ 300 / 1.414 ≈ 212.1 V.

(b) To find the peak current (I_m) from the rms current (I_rms):
The formula is I_rms = I_m / √2, which can be rearranged to I_m = I_rms * √2.
Given I_rms = 10 A.
I_m = 10 * √2 ≈ 10 * 1.414 = 14.14 A.

Question 3

Q3. A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Accepted Answer

To find the RMS current in a purely inductive circuit, we first need to calculate the inductive reactance (X_L), which is the opposition offered by the inductor to the flow of AC. Then, we can use Ohm's law for AC circuits.

Given:
Inductance, L = 44 mH = 44 × 10⁻³ H
RMS voltage, V_rms = 220 V
Frequency, f = 50 Hz

Step 1: Calculate the inductive reactance (X_L).
The formula for inductive reactance is X_L = 2πfL.
X_L = 2 * 3.14 * 50 Hz * (44 × 10⁻³ H)
X_L = 100 * 3.14 * 44 × 10⁻³ Ω
X_L = 13.816 Ω ≈ 13.82 Ω

Step 2: Calculate the RMS current (I_rms).
Using Ohm's law for an AC circuit, I_rms = V_rms / X_L.
I_rms = 220 V / 13.82 Ω
I_rms ≈ 15.92 A

Question 4

Q4. A 60 µF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Accepted Answer

To find the RMS current in a purely capacitive circuit, we first need to calculate the capacitive reactance (X_C), which is the opposition offered by the capacitor to the flow of AC. Then, we can use Ohm's law for AC circuits.

Given:
Capacitance, C = 60 µF = 60 × 10⁻⁶ F
RMS voltage, V_rms = 110 V
Frequency, f = 60 Hz

Step 1: Calculate the capacitive reactance (X_C).
The formula for capacitive reactance is X_C = 1 / (2πfC).
X_C = 1 / (2 * 3.14 * 60 Hz * 60 × 10⁻⁶ F)
X_C = 1 / (22608 × 10⁻⁶) Ω
X_C ≈ 44.23 Ω

Step 2: Calculate the RMS current (I_rms).
Using Ohm's law for an AC circuit, I_rms = V_rms / X_C.
I_rms = 110 V / 44.23 Ω
I_rms ≈ 2.49 A

Question 5

Q5. In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Accepted Answer

In both cases, the net power absorbed by each circuit over a complete cycle is zero.

**Explanation:**

The average power (P) consumed in an AC circuit is given by the formula:
P = V_rms * I_rms * cos(φ)
where φ is the phase angle between the voltage and the current, and cos(φ) is the power factor.

1.  **For the Purely Inductive Circuit (Exercise 7.3):**
    In a circuit containing only an ideal inductor, the current lags behind the voltage by a phase angle of π/2 radians or 90°. 
    Therefore, φ = 90°.
    The power factor is cos(φ) = cos(90°) = 0.
    So, the net power absorbed is P = V_rms * I_rms * 0 = 0.
    Physically, during one quarter of the cycle, the inductor stores energy in its magnetic field from the source. In the next quarter cycle, it returns this stored energy completely to the source. This process repeats, so over a full cycle, there is no net consumption of energy.

2.  **For the Purely Capacitive Circuit (Exercise 7.4):**
    In a circuit containing only an ideal capacitor, the current leads the voltage by a phase angle of π/2 radians or 90°.
    Therefore, φ = 90°.
    The power factor is cos(φ) = cos(90°) = 0.
    So, the net power absorbed is P = V_rms * I_rms * 0 = 0.
    Physically, during one quarter of the cycle, the capacitor stores energy in its electric field as it charges. In the next quarter cycle, it returns this stored energy completely to the source as it discharges. Over a full cycle, the net energy consumption is zero.

The current in such purely reactive (inductive or capacitive) circuits is known as **wattless current** because it consumes no power.

Question 6

Q6. A 100 µF capacitor in series with a 40 Ω resistor is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the curr

Accepted Answer

The circuit consists of a capacitor and a resistor in series (RC circuit). We first calculate the capacitive reactance, then the impedance of the circuit. From the impedance and RMS voltage, we find the maximum current. Finally, we determine the phase angle to calculate the time lag between current and voltage maxima.

Question 7

Q7. An LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which the current amplitude is maximum

Accepted Answer

This question explores various aspects of a series LCR circuit, including resonance frequency, maximum current and power, half-power frequencies, and the Q-factor. Resonance occurs when the inductive and capacitive reactances cancel each other out, leading to maximum current and power.

Question 8

Q8. A series LCR circuit with L = 5.0 H, C = 80 µF, R = 40 Ω is connected to a variable frequency 230 V supply.
(a) Determine the source frequency which drives the circuit in resonanc

Accepted Answer

This problem deals with a series LCR circuit at resonance. We will calculate the resonant frequency, impedance, current amplitude, and RMS voltage drops across each component. A key property of resonance is that the voltage drops across the inductor and capacitor are equal and opposite, leading to a zero net voltage drop across the LC combination.

Question 9

Q9. A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of t

Accepted Answer

When the frequency of the supply equals the natural frequency of the circuit, the circuit is in resonance. At resonance, the impedance of the circuit is purely resistive (Z = R), and the phase angle (φ) between the voltage and current is zero. The average power transferred to the circuit in one complete cycle is given by P_avg = V_rms * I_rms * cos(φ). Since cos(φ) = cos(0) = 1 at resonance, P_avg = V_rms * I_rms. Also, at resonance, I_rms = V_rms / R.

Given:
Resistance, R = 20 Ω
Inductance, L = 1.5 H
Capacitance, C = 35 μF = 35 × 10⁻⁶ F
RMS voltage of the supply, V_rms = 200 V

At resonance:
Impedance, Z = R = 20 Ω
RMS current, I_rms = V_rms / R = 200 V / 20 Ω = 10 A

Average power transferred, P_avg = V_rms * I_rms * cos(φ)
Since φ = 0 at resonance, cos(φ) = 1.
P_avg = 200 V × 10 A × 1 = 2000 W

Alternatively, P_avg = I_rms² * R = (10 A)² × 20 Ω = 100 A² × 20 Ω = 2000 W.

The average power transferred to the circuit in one complete cycle at resonance is 2000 W.

Question 10

Q10. An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be

Accepted Answer

Given:
Inductance, L = 20 mH = 20 × 10⁻³ H
Capacitance, C = 50 μF = 50 × 10⁻⁶ F
Initial charge on capacitor, q₀ = 10 mC = 10 × 10⁻³ C
Resistance R = 0 (negligible)

**(a) Total energy stored initially and its conservation:**
At t = 0, the capacitor has an initial charge q₀, and the circuit is just closed, so the current in the inductor is zero. Therefore, all the energy stored initially is electrical energy in the capacitor.
Initial electrical energy, U_E = q₀² / (2C)
U_E = (10 × 10⁻³ C)² / (2 × 50 × 10⁻⁶ F)
U_E = (100 × 10⁻⁶) / (100 × 10⁻⁶) J
U_E = 1 J

Since the resistance of the circuit is negligible, there is no energy dissipation due to resistance. Therefore, the total energy stored in the LC circuit (sum of electrical energy in the capacitor and magnetic energy in the inductor) is conserved during LC oscillations. The energy continuously oscillates between the capacitor (electrical) and the inductor (magnetic).

**(b) Natural frequency of the circuit:**
The natural angular frequency (ω₀) of an LC circuit is given by:
ω₀ = 1 / √(LC)
ω₀ = 1 / √((20 × 10⁻³ H) × (50 × 10⁻⁶ F))
ω₀ = 1 / √(1000 × 10⁻⁹)
ω₀ = 1 / √(10⁻⁶)
ω₀ = 1 / (10⁻³) = 1000 rad/s

The natural frequency (ν₀) is given by:
ν₀ = ω₀ / (2π)
ν₀ = 1000 / (2π) Hz ≈ 159.24 Hz

**(c) Times for completely electrical or completely magnetic energy:**
The charge on the capacitor in an LC circuit undergoing oscillations is given by q(t) = q₀ cos(ω₀t), assuming initial charge q₀ and zero initial current.

(i) **Completely electrical energy:** Energy is completely electrical when the charge on the capacitor is maximum (q = ±q₀) and the current in the inductor is zero. This occurs when cos(ω₀t) = ±1.
ω₀t = nπ, where n = 0, 1, 2, 3, ...
t = nπ / ω₀
t = nπ / 1000 s
For n = 0, t = 0 s (initial condition)
For n = 1, t = π / 1000 s ≈ 3.14 × 10⁻³ s
For n = 2, t = 2π / 1000 s ≈ 6.28 × 10⁻³ s
And so on. So, the energy is completely electrical at t = 0, π/ω₀, 2π/ω₀, ...

(ii) **Completely magnetic energy:** Energy is completely magnetic when the charge on the capacitor is zero (q = 0) and the current in the inductor is maximum. This occurs when cos(ω₀t) = 0.
ω₀t = (n + 1/2)π, where n = 0, 1, 2, 3, ...
t = (n + 1/2)π / ω₀
t = (2n + 1)π / (2ω₀)
t = (2n + 1)π / (2 × 1000) s
For n = 0, t = π / (2 × 1000) s ≈ 1.57 × 10⁻³ s
For n = 1, t = 3π / (2 × 1000) s ≈ 4.71 × 10⁻³ s
And so on. So, the energy is completely magnetic at t = π/(2ω₀), 3π/(2ω₀), 5π/(2ω₀), ...

**(d) Times when total energy is shared equally:**
When the total energy (U) is shared equally, the electrical energy (U_E) equals the magnetic energy (U_B), and each is half of the total energy.
U_E = U_B = U / 2
Since U_E = q² / (2C) and U = q₀² / (2C), we have:
q² / (2C) = (1/2) * (q₀² / (2C))
q² = q₀² / 2
q = ± q₀ / √2

Using q(t) = q₀ cos(ω₀t):
q₀ cos(ω₀t) = ± q₀ / √2
cos(ω₀t) = ± 1 / √2

This occurs when ω₀t = π/4, 3π/4, 5π/4, 7π/4, ... (or generally (2n+1)π/4 for n=0,1,2,3... and (2n-1)π/4 for n=1,2,3...)
So, ω₀t = (2n + 1)π/4 for n = 0, 1, 2, 3, ...
t = (2n + 1)π / (4ω₀)
t = (2n + 1)π / (4 × 1000) s

For n = 0, t = π / (4 × 1000) s ≈ 0.785 × 10⁻³ s
For n = 1, t = 3π / (4 × 1000) s ≈ 2.356 × 10⁻³ s
For n = 2, t = 5π / (4 × 1000) s ≈ 3.927 × 10⁻³ s
And so on. So, the energy is shared equally at t = π/(4ω₀), 3π/(4ω₀), 5π/(4ω₀), ...

Question 11

Q11. A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maxim

Accepted Answer

Given:
Inductance, L = 0.12 H
Capacitance, C = 480 nF = 480 × 10⁻⁹ F
Resistance, R = 23 Ω
RMS voltage of the supply, V_rms = 230 V

**(a) Source frequency for maximum current amplitude and maximum current amplitude:**
The current amplitude is maximum when the circuit is in resonance. At resonance, the source frequency (ν₀) is the natural frequency of the circuit.
Angular resonant frequency, ω₀ = 1 / √(LC)
ω₀ = 1 / √((0.12 H) × (480 × 10⁻⁹ F))
ω₀ = 1 / √(57.6 × 10⁻⁹)
ω₀ = 1 / √(576 × 10⁻¹⁰)
ω₀ = 1 / (24 × 10⁻⁵) = 10⁵ / 24 ≈ 4166.67 rad/s

Resonant frequency, ν₀ = ω₀ / (2π)
ν₀ = 4166.67 / (2π) ≈ 663.15 Hz

At resonance, the impedance Z = R = 23 Ω.
The RMS current at resonance, I_rms = V_rms / R = 230 V / 23 Ω = 10 A.
The maximum current amplitude (I_m) is related to the RMS current by I_m = I_rms√2.
I_m = 10√2 A ≈ 14.14 A

**(b) Source frequency for maximum average power and maximum average power:**
The average power absorbed by the circuit is maximum at resonance, which occurs at the same frequency as for maximum current amplitude.
So, the source frequency for maximum average power is ν₀ ≈ 663.15 Hz.

Maximum average power, P_max = V_rms * I_rms * cos(φ)
At resonance, φ = 0, so cos(φ) = 1.
P_max = 230 V × 10 A × 1 = 2300 W
Alternatively, P_max = I_rms² * R = (10 A)² × 23 Ω = 100 × 23 W = 2300 W.

**(c) Half-power frequencies:**
The power transferred to the circuit is half the power at resonant frequency (P_max/2) when the current amplitude is I_m / √2 or the RMS current is I_rms / √2. This condition occurs when the impedance Z = R√2, which implies |X_L - X_C| = R.
Let the half-power frequencies be ω₁ and ω₂. The bandwidth is Δω = ω₂ - ω₁ = R/L.
We know ω₀ = 4166.67 rad/s.
Δω = R/L = 23 Ω / 0.12 H ≈ 191.67 rad/s

The half-power frequencies are approximately given by:
ω₁ = ω₀ - Δω/2 = 4166.67 - 191.67/2 = 4166.67 - 95.835 ≈ 4070.835 rad/s
ω₂ = ω₀ + Δω/2 = 4166.67 + 191.67/2 = 4166.67 + 95.835 ≈ 4262.505 rad/s

Converting to frequencies (in Hz):
ν₁ = ω₁ / (2π) = 4070.835 / (2π) ≈ 647.9 Hz
ν₂ = ω₂ / (2π) = 4262.505 / (2π) ≈ 678.5 Hz

**(d) Q-factor of the circuit:**
The Q-factor (Quality factor) of a series LCR circuit is given by:
Q = (ω₀L) / R or Q = 1 / (ω₀CR) or Q = (1/R)√(L/C)

Using Q = (ω₀L) / R:
Q = (4166.67 rad/s × 0.12 H) / 23 Ω
Q = 500 / 23 ≈ 21.74

Using Q = 1 / (ω₀CR):
Q = 1 / (4166.67 rad/s × 480 × 10⁻⁹ F × 23 Ω)
Q = 1 / (0.046) ≈ 21.74

Using Q = (1/R)√(L/C):
Q = (1/23)√((0.12 H) / (480 × 10⁻⁹ F))
Q = (1/23)√(0.25 × 10⁶)
Q = (1/23) × (0.5 × 10³) = 500 / 23 ≈ 21.74

Question 12

Q12. A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of t

Accepted Answer

When the frequency of the supply equals the natural frequency of the circuit, the circuit is in resonance. At resonance, the impedance of the circuit is purely resistive (Z = R), and the current in the circuit is maximum. The average power transferred to the circuit in one complete cycle at resonance is 2000 W.

Question 13

Q13. An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be

Accepted Answer

(a) The total energy stored initially is purely electrical, stored in the capacitor. It is 1 J. Yes, in an ideal LC circuit with negligible resistance, the total energy (sum of electrical and magnetic energy) is conserved during LC oscillations.
(b) The natural frequency of the circuit is approximately 159.15 Hz.
(c) (i) The energy is completely electrical at t = 0, T, 2T, ... (i.e., at t = nT, where n = 0, 1, 2, ...).
    (ii) The energy is completely magnetic at t = T/4, 3T/4, 5T/4, ... (i.e., at t = (2n+1)T/4, where n = 0, 1, 2, ...).
(d) The total energy is shared equally between the inductor and the capacitor at t = T/8, 3T/8, 5T/8, 7T/8, ... (i.e., at t = (2n+1)T/8, where n = 0, 1, 2, ...).

Question 14

Q14. A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Accepted Answer

The angular frequency of free oscillations of the LC circuit is approximately 1111.11 rad/s.

Question 15

Q15. An L-C-R series circuit with 100 Ω resistor is connected to an AC source of 200 V and angular frequency 300 rad/s. When only the capacitor is removed, the current lags the voltage

Accepted Answer

Given:
Resistance, R = 100 Ω
RMS voltage of AC source, V_rms = 200 V
Angular frequency, ω = 300 rad/s

Case 1: Only the capacitor is removed (RL circuit).
The current lags the voltage by 60°.
Phase angle, φ₁ = +60° (lagging current implies X_L > X_C, and here X_C = 0, so X_L > 0)
tan φ₁ = X_L / R
tan 60° = X_L / 100
√3 = X_L / 100
X_L = 100√3 Ω

Case 2: Only the inductor is removed (RC circuit).
The current leads the voltage by 60°.
Phase angle, φ₂ = -60° (leading current implies X_C > X_L, and here X_L = 0, so X_C > 0)
tan φ₂ = -X_C / R
tan (-60°) = -X_C / 100
-√3 = -X_C / 100
X_C = 100√3 Ω

For the complete L-C-R circuit:
We have X_L = 100√3 Ω and X_C = 100√3 Ω.
Since X_L = X_C, the circuit is in resonance.

Impedance of the L-C-R circuit at resonance, Z = R.
Z = 100 Ω

Current in the circuit (RMS value):
I_rms = V_rms / Z
I_rms = 200 V / 100 Ω
I_rms = 2 A

Average power dissipated in the L-C-R circuit:
At resonance, the phase angle φ = 0° (since X_L = X_C).
Power factor, cos φ = cos 0° = 1.
P_avg = V_rms * I_rms * cos φ
P_avg = 200 V * 2 A * 1
P_avg = 400 W

Alternatively, P_avg = I_rms² * R
P_avg = (2 A)² * 100 Ω
P_avg = 4 * 100 W
P_avg = 400 W

Therefore, the current in the L-C-R circuit is 2 A and the average power dissipated is 400 W.

Question 16

Q16. An L-C-R series circuit with L = 3.0 H, C = 27 μF, R = 7.4 Ω is connected to a variable frequency 200 V AC source.
(a) Determine the angular frequency of the source at which the c

Accepted Answer

Given:
Inductance, L = 3.0 H
Capacitance, C = 27 μF = 27 × 10⁻⁶ F
Resistance, R = 7.4 Ω
RMS voltage of AC source, V_rms = 200 V

**(a) Angular frequency of resonance (ω₀):**
At resonance, the angular frequency ω₀ is given by:
ω₀ = 1 / √(LC)
ω₀ = 1 / √((3.0 H) × (27 × 10⁻⁶ F))
ω₀ = 1 / √(81 × 10⁻⁶)
ω₀ = 1 / (9 × 10⁻³)
ω₀ = 1000 / 9 rad/s
ω₀ ≈ 111.11 rad/s

**(b) Impedance of the circuit and amplitude of current at resonance:**
At resonance, the impedance Z is purely resistive, i.e., Z = R.
Z = 7.4 Ω

To find the amplitude of current (I₀), we first need the peak voltage (V₀).
V_rms = V₀ / √2
V₀ = V_rms × √2
V₀ = 200 V × √2 ≈ 200 × 1.414 = 282.8 V

Amplitude of current, I₀ = V₀ / Z
I₀ = 282.8 V / 7.4 Ω
I₀ ≈ 38.22 A

**(c) RMS potential drop across the three elements and potential drop across L-C combination:**
First, calculate the RMS current at resonance:
I_rms = V_rms / Z = V_rms / R
I_rms = 200 V / 7.4 Ω
I_rms ≈ 27.03 A

Now, calculate the RMS potential drop across each element:
Potential drop across the resistor, V_R = I_rms × R
V_R = 27.03 A × 7.4 Ω
V_R ≈ 200 V
(This is expected, as at resonance, the entire source voltage drops across the resistor).

Inductive reactance at resonance, X_L = ω₀L
X_L = (1000/9 rad/s) × 3.0 H
X_L = 1000/3 Ω ≈ 333.33 Ω

Capacitive reactance at resonance, X_C = 1 / (ω₀C)
X_C = 1 / ((1000/9 rad/s) × (27 × 10⁻⁶ F))
X_C = 1 / (3 × 10⁻³)
X_C = 1000/3 Ω ≈ 333.33 Ω
(As expected, X_L = X_C at resonance).

Potential drop across the inductor, V_L = I_rms × X_L
V_L = 27.03 A × (1000/3 Ω)
V_L ≈ 9010 V

Potential drop across the capacitor, V_C = I_rms × X_C
V_C = 27.03 A × (1000/3 Ω)
V_C ≈ 9010 V

Potential drop across the L-C combination:
The potential drop across the L-C combination is the vector sum of V_L and V_C. Since V_L and V_C are 180° out of phase, their resultant is V_L - V_C.
Potential drop across L-C combination = V_L - V_C
= 9010 V - 9010 V
= 0 V

This shows that the potential drop across the L-C combination is zero at the resonating frequency, as V_L and V_C are equal in magnitude and opposite in phase.

Question 17

Q17. A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Accepted Answer

Given:
Capacitance, C = 30 μF = 30 × 10⁻⁶ F
Inductance, L = 27 mH = 27 × 10⁻³ H

The angular frequency (ω) of free oscillations in an LC circuit is given by the formula:
ω = 1 / √(LC)

Substitute the given values:
ω = 1 / √((27 × 10⁻³ H) × (30 × 10⁻⁶ F))
ω = 1 / √(810 × 10⁻⁹)
ω = 1 / √(81 × 10⁻⁸)
ω = 1 / (9 × 10⁻⁴)
ω = 10⁴ / 9 rad/s
ω ≈ 1111.11 rad/s

Therefore, the angular frequency of free oscillations of the circuit is approximately 1111.11 rad/s.

Chapter 7: Alternating Current