NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current - Param Himalaya
7.1. 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?
Solution :
(a) $I_{rms} = \frac{v_{rms}}{R} = \frac{220}{100} = 2.20 A$
(b) $Net power = V_{rms} \times I_{rms} = 220 \times 2.20$
= 484 W
7.2 (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?
Solution:
(a) $V_{rms} = \frac{V_{0}}{\sqrt{2}} =\frac{300}{\sqrt{2}} = 212.1 V$
(b) $I_{rms} = \frac{I_{0}}{\sqrt{2}}$
$I_{0} = I_{rms} \sqrt{2} = 10 \sqrt{2} = 14.1A$
7.3 A 44 mH inductor is connected to 220 V , 50 Hz ac supply. Determine the rms value of the current in the circuit.
Solution:
Here , Reactance $X_{L} = 2 \pi\nu L = 2\pi \times 50 \times 44 \times 10^{-3}$
$\therefore I_{rms} = \frac{V_{rms}}{X_{L}} = \frac{220}{2\pi\times 50 \times 44 \times 10^{-3}}$
$I_{rms} = 15.91 A$
7.4 A 60 µF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Solution :
Here , Reactance $X_{L} = \frac{1}{2 \pi\nu C} = \frac{1}{2\pi \times 60 \times 60 \times 10^{-6}}$
$\therefore I_{rms} = \frac{V_{rms}}{X_{C}} = \frac{110}{\frac{1}{2\pi \times 60 \times 60 \times 10^{-6}}}$
$I_{rms} = 110 \times 2\pi \times 60 \times 60 \times 10^{-6}$
$I_{rms} = 2.49A$
7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each. circuit over a complete cycle. Explain your answer.
Solution :
In the inductive circuit,R
Rms value of current, I = 15.92 A
Rms value of voltage, V = 220 V
Hence, the net power absorbed can be obtained by the relation,
P = VI cos Φ
Where, Φ = Phase difference between V and I For a pure inductive circuit,
the phase difference between alternating voltage and current is 90°
i.e., Φ= 90°.
P = 220 × 15.92 × cos 90° Hence,
P = 0
i.e., the net power is zero.
In the capacitive circuit,
Rms value of current, I = 2.49 A
Rms value of voltage, V = 110 V
Hence, the net power absorbed can ve obtained as:
P = VI Cos Φ
For a pure capacitive circuit, the phase difference between alternating voltage and current is 90°
i.e., Φ= 90°.
P = 110 × 2.49 × cos 90°
Hence, P = 0 i.e., the net power is zero.
7.6 A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Solution :
7.7 A series LCR circuit with R = 200, 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 the circuit, what is the average power transferred to the circuit in one complete cycle?
Solution :
7.8 Figure 7.17 shows a series LCR circuit connected to a variable. frequency 230 V source. L 5.0 H, C 80µF, R = 40 Ω.
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Solution :
(a) Resonant angular frequency :
$$\omega = \frac{1}{\sqrt{LC}}$$
$$\omega = \frac{1}{\sqrt{5\times80\times10^{-6}}$$
$$\omega = \frac{1}{\sqrt{5\times80\times10^{-6}}}$$
$$\omega = 50 rad s^{-1}$$
(b) Circuit impedance,
$$Z = \sqrt{R^{2}+( \omega L - \frac{1}{\omega C})^{2}}$$
$$\omega L = \frac{1}{\omega C}$$
So that
$$Z= \sqrt{R^{2}}$$
$$Z= R = 40 \Omega$$
Amplitude of current at resonant frequency is
$$I_{0} = \frac{E_{O}}{Z}$$
$$I_{0} = \frac{\sqrt2 E_{v}}{R}$$
$$= \frac{\sqrt2 \times 230}{40}$$
$$I_{0}= 8.13 A$$
(c)$$ I_{v} = \frac{E_{v}}{R} = \frac{230}{40}$$
$$I_{v} = 5.75 A$$
P.D across L,
$$V_{L} = I_{v} \times \omega_{r}L$$
$$V_{L} = 5.75 \times 50 \times 5.0$$
$$V_{L} = 1437.5 v$$
P.D across C ,
$$V_{C} = I_{v} \times \frac{1}{\omega_{r}C}$$
$$V_{C} = 5.75 \times \frac{1}{50 \times 80 \times 10^{-6}$$
$$V_{C} = 1437.5 V$$
P.D across R ,
$$V_{R} = I_{v}R$$
$$V_{R} = I_{v}R$$
$$V_{R} = 5.75 \times 40 = 230 V$$
P.D drop across L-C combination
$$V_{LC} = I_{v} ( \omega_{r}L - \frac{1}{\omega_{r}C})$$
$$V_{LC} = I_{v}(0) = 0$$