DC Pandey Solutions: Alternating Current | Physics Class 12 - NEET PDF Download

Introductory Exercise 25.1

Ques 1: An electric lamp which runs at 100 V dc and consumes 10 A current is connected to ac mains at 150 V, 50 Hz cycles with a choke coil in series. Calculate the inductance and drop of voltage across the choke. Neglect the resistance of choke.
Sol: When the lamp runs at 100 V dc and 10 A, its effective resistance is R = 100/10 = 10 Ω.
On ac mains of V = 150 V (rms) and desired current I = 10 A, the inductive reactance X_L must satisfy:
V = I·√(R² + X_L²) 
⇒ X_L = √((V/I)² - R²) 
= √((150/10)² - 10²) 
= √(225 - 100) 
= 11.18 Ω.
At f = 50 Hz, L = X_L / (2πf) = 11.18 / (2π × 50) = 0.0356 H ≈ 0.036 H.
Voltage drop across choke V_L = I·X_L = 10 × 11.18 = 111.8 V.

Ques 2: A circuit operating at  contains a 1μF capacitor and a 20Ωresistor. How large an inductor must be added in series to make the phase angle for the circuit zero? Calculate the current in the circuit if the applied voltage is 120 V.
Sol: Resonance (phase angle zero) requires X_L = X_C.

For C = 1.00 μF = 1.00 × 10^-6 F and f = 50 Hz, ω = 2πf = 314.16 rad/s.

X_C = 1/(ωC) 
= 1/(314.16 × 1.00×10^-6) 
= 3183.1 Ω.

Thus L = X_L/ω 
= X_C/ω 
= 3183.1 / 314.16 
= 10.13 H.
At resonance the net reactance is zero, so circuit impedance  R = 20 Ω and current I = V/R = 120/20 = 6.0 A.

Introductory Exercise 25.2

Ques 1: If a 0.03 H inductor, a 10Ω resistor and a 2μF capacitor are connected in series. At what frequency will they resonate? What will be the phase angle at resonance?
Sol:

 Ques 2: An arc lamp consumes 10 A at 40 V. Calculate the power factor when it is connected with a suitable value of choke coil required to run the arc lamp on ac mains of 200 V (rms) and 50 Hz.
Sol:

= 0.2

Exercises
For JEE Main

Subjective Questions
Ques 1: (a) What is the reactance of a 2.00 H inductor at a frequency of 50.0 Hz? 
(b) What is the inductance of an inductor whose reactance is 2.00Ω at 50.0 Hz? 
(c) What is the reactance of a 2.00 μF capacitor at a frequency of 50.0 Hz? 
(d) What is the capacitance of a capacitor whose reactance is 2.00Ω at 50.0 Hz?
Sol: 

(a) X_L = 2π f L = 2π × 50.0 × 2.00 = 628.32 Ω.

(b) L = X_L/(2π f) = 2.00 / (2π × 50.0) = 2.00 / 314.16 = 0.006366 H = 6.366 mH.

(c) X_C = 1 / (2π f C) = 1 / (2π × 50.0 × 2.00×10^-6) = 1591.55 Ω.

(d) C = 1 / (2π f X_C) = 1 / (2π × 50.0 × 2.00) = 1 / 628.32 = 1.5915×10^-3 F = 1591.5 μF.

Ques 2: A 300 Ω resistor, a 0.250 H inductor, and a 8.00 μF capacitor are in series with an ac source with voltage amplitude 120 V and angular frequency 400 rad/s. 
(a) What is the current amplitude? 
(b) What is the phase angle of the source voltage with respect to the current? Does the source voltage lag or lead the current? 
(c) What are the voltage amplitudes across the resistor, inductor, and capacitor?
Sol: X_L = ωL = 100Ω

(b) Since X_C > X_L, voltage lags the current by an angle given by

(c) (V₀)_R = I₀R = (0.326) 300 = 97.8 V
(V₀)_L = I₀X_L = (0.326) (100) = 32.6 V
(V₀)_C = I₀X_C = (0.326) (312.5) = 102 V
Complete numerical solution and explanation:
X_L = ωL = 400 × 0.250 = 100 Ω.
X_C = 1/(ωC) = 1 / (400 × 8.00×10^-6) = 312.5 Ω.
Net reactance X = X_L - X_C = 100 - 312.5 = -212.5 Ω.
Impedance magnitude Z = √(R² + X²) = √(300² + (-212.5)²) = 367.6 Ω.
Current amplitude I₀ = V₀ / Z = 120 / 367.6 = 0.3265 A (≈ 0.326 A).
Phase angle φ = arctan(X / R) = arctan(-212.5 / 300) = -35.3° (negative ⇒ voltage lags current by 35.3°).
Voltage amplitudes: V_R0 = I₀R = 0.3265 × 300 = 97.95 V.
V_L0 = I₀X_L = 0.3265 × 100 = 32.65 V.
V_C0 = I₀X_C = 0.3265 × 312.5 = 102.03 V.

Ques 3: In an L-C-R series circuit, R = 150Ω, L = 0.750 H, and C = 0.0180 μF. The source has voltage amplitude V = 150 V and a frequency equal to the resonance frequency of the circuit.
(a) What is the power factor?
(b) What is the average power delivered by the source?
(c) The capacitor is replaced by one wit h C = 0.0360 μB and the source frequency is adjusted to the new resonance value. Then, what is the average power delivered by the source?
Sol:  (a) At resonance the circuit is purely resistive, so power factor = 1.0.

(b) Given source amplitude V₀ = 150 V 
⇒ V_rms = V₀/√2 
= 150/√2 
= 106.07 V.

Average power P = V_rms² / R 
= (106.07)² / 150 
= 11250 / 150 
= 75.0 W.

(c) Replacing C and adjusting frequency so circuit is again at resonance does not change R or V. At resonance Z = R, so average power remains P = 75.0 W.

Ques 4: A series circuit has an impedance of 60.0 W and a power factor of 0.720 at 50.0 Hz. The source voltage lags the current.
(a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor?
(b) What size element will raise the power factor to unit y?
Sol: (a) Voltage lags the current, which means current leads voltage and the circuit is net capacitive (X_C > X_L). To reduce the net capacitive reactance and improve the power factor, an inductor should be added in series.

To increase the power factor the denominator in cosφ = R/Z should decrease. Adding inductance will increase X_L and cancel part of the capacitive reactance.

Hence X_L should increase. Therefore an inductor is required to be connected.

Compute required value: R = pf × Z = 0.72 × 60.0 = 43.2 Ω.

Current net reactive magnitude X = √(Z² - R²) = √(60² - 43.2²) 
= √(3600 - 1866.24) 
= √1733.76 = 41.64 Ω. (This is the magnitude of net capacitive reactance to be cancelled.)
Required inductive reactance X_L,new = 41.64 Ω to bring net reactance to zero.
At f = 50.0 Hz, 
L = X_L / (2πf) = 41.64 / (2π × 50) 
= 41.64 / 314.16 
= 0.1326 H 
≈ 0.133 H.

Ques 5: Voltage and current for a circuit with two elements in series are expressed as :
V(t) = 170sin (6280 t +π/3) volt
i(t) = 8.5sin(6280t + π/2)amp
(a) Plot the two waveforms. 
(b) Determine the frequency in Hz. 
(c) Determine the power factor stating its nature. 
(d) What are the values of the elements?
Sol: 

Given: V(t) = 170 sin(6280 t + π/3) volt  

i(t) = 8.5 sin(6280 t + π/2) amp

(a) Plot the waveforms

Both voltage and current are sinusoidal and have the same angular frequency.
Phase of voltage = π/3 = 60°
Phase of current = π/2 = 90°
Therefore, current leads voltage by:
φ = π/2 − π/3 = π/6 = 30°
Hence, the current waveform leads the voltage waveform by 30°.

(b) Frequency of the supply
Angular frequency: ω = 6280 rad/s
Frequency:
f = ω / (2π)
f = 6280 / (2π)
f ≈ 1000 Hz

(c) Power factor and nature of circuit
Phase difference: φ = (π/3 − π/2) = −π/6 = −30°
Power factor: cos φ = cos 30°
cos φ = √3 / 2 ≈ 0.866
Since current leads voltage, the power factor is leading.
Power factor = 0.866 (leading)

(d) Values of the circuit elements
Peak voltage: V₀ = 170 V
Peak current: I₀ = 8.5 A

Impedance:
Z = V₀ / I₀
Z = 170 / 8.5
Z = 20 Ω
Resistance:
R = Z cos φ
R = 20 × cos 30°
R = 20 × 0.866
R = 17.32 Ω
Reactance:
X = Z sin φ
X = 20 × sin(−30°)
X = −10 Ω
Negative sign indicates capacitive reactance.

Capacitance: Xc = 1 / (ωC)
C = 1 / (ω Xc)
C = 1 / (6280 × 10)
C = 1.592 × 10⁻⁵ F
C = 15.92 μF

Final Answer:

Frequency = 1000 Hz  , Power factor = 0.866 (leading)  
Circuit elements:  Resistance R = 17.32 Ω  , Capacitance C = 15.92 μF  
The circuit is a series R–C circuit.

Ques 6: A 5.00 H inductor with negligible resistance is connected across an ac source. Voltage amplitude is kept constant at 60.0 V but whose frequency can be varied. Find the current amplitude when the angular, frequency is
(a) 100 rad/s (b) 1000 rad/s (c) 10000 rad/s
Sol: 

Given L = 5.00 H and V₀ = 60.0 V.

(a) ω = 100 rad/s ⇒ X_L = 100 × 5.00 = 500 Ω ⇒ I₀ = 60.0 / 500 = 0.120 A.

(b) ω = 1000 rad/s ⇒ X_L = 1000 × 5.00 = 5000 Ω ⇒ I₀ = 60.0 / 5000 = 0.0120 A.

(c) ω = 10000 rad/s ⇒ X_L = 10000 × 5.00 = 50000 Ω ⇒ I₀ = 60.0 / 50000 = 0.00120 A.

Ques 7: A 300Ω resistor is connected in series wit h a 0.800 H inductor. The voltage across the resistor as a function of time is V_R = (2.50 V)cos [(950rad/s)t].
(a) Derive an expression for the circuit current.
(b) Determine the inductive reactance of the inductor.
(c) Derive an expression for the voltage V_L across the inductor.
Sol: 

= 8.33 mA
Current function and V_R function are in phase. Hence,

I = (8.33 mA) cos [(950 rad/s) t]

(b) X_L= ωL = 950 × 0.8 = 760Ω

(c) (V₀)_L = I₀X_L = (8.33 × 10^-3) (760)

= 6.33V
Now, V_L function leads the current (or V_R ) function by 90°.
V_L(t) = 6.33 cos(950 t + 90°) = -6.33 sin (950 t).
Detailed steps:
Amplitude across R is V_R0 = 2.50 V. Hence I₀ = V_R0 / R = 2.50 / 300 = 8.333×10^-3 A = 8.33 mA.
Current in time domain i(t) = I₀ cos(950 t) (in phase with V_R).
X_L = ωL = 950 × 0.8 = 760 Ω.
V_L0 = I₀ X_L = 8.333×10^-3 × 760 = 6.333 V and V_L leads by 90°.

Ques 8: An L-C-R series circuit with L = 0.120 H, R = 240Ω, and C = 7.30 μF carries an rms current of 0.450 A with a frequency of 400 Hz.
(a) What are the phase angle and power factor for this circuit?
(b) What is the impedance of the circuit? 
(c) What is the rms voltage of the source? 
(d) What average power is delivered by the source? 
(e) What is the average rate at which electrical energy is converted to thermal energy in the resistor? 
(f) What is the average rate at which electrical energy is dissipated ( converted to other forms) in the capacitor? 
(g) In the inductor?
Sol: Given L = 0.120 H, R = 240 Ω, C = 7.30 μF = 7.30×10^-6 F, f = 400 Hz.
ω = 2π f = 2π × 400 = 2513.27 rad/s.
X_L = ωL = 2513.27 × 0.120 = 301.59 Ω (≈ 302 Ω).
X_C = 1 / (ωC) = 1 / (2513.27 × 7.30×10^-6) = 54.54 Ω.
Net reactance X = X_L - X_C = 301.59 - 54.54 = 247.05 Ω.
(a) Phase angle φ = arctan(X / R) = arctan(247.05 / 240) = 45.66° (voltage leads current because X_L > X_C).
Power factor = cos φ = cos 45.66° = 0.699 (approximately).
(b) Impedance Z = √(R² + X²) = √(240² + 247.05²) = 344.46 Ω (≈ 344 Ω).

(c) Source rms voltage V_rms = I_rms × Z = 0.450 × 344.46 = 155.0 V (approximately).

(d) Average power delivered by the source P = I_rms² R = (0.450)² × 240 = 0.2025 × 240 = 48.6 W.

(e) Rate of conversion to thermal energy in resistor = P_R = 48.6 W.

(f) Average power for ideal capacitor P_C = 0 (no net energy dissipated over a full cycle).

(g) Average power for ideal inductor P_L = 0 (no net energy dissipated over a full cycle).