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Alternating Current

CBSE · Class 12 · Physics

NCERT Solutions for Alternating Current — CBSE Class 12 Physics.

110 questions76 flashcards5 concepts

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A graph showing the instantaneous voltage and current waveforms in a purely resistive AC circuit, demonstrating that they are in phase with each other.
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8 Questions Solved · 1 Section

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EXERCISES

7.1A 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?Show solution
For a pure resistor, Ohm's law applies to rms values:

I=VR=220100=2.2 AI = \frac{V}{R} = \frac{220}{100} = 2.2\,\text{A}

For a resistor, the power consumed over a full cycle is

P=IV=(2.2)(220)=484 WP = IV = (2.2)(220) = 484\,\text{W}

So the rms current is 2.2 A2.2\,\text{A} and the net power consumed is 484 W484\,\text{W}.

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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?Show solution
(a) For ac quantities,

V=vm2=3002≈212.1 VV = \frac{v_m}{\sqrt{2}} = \frac{300}{\sqrt{2}} \approx 212.1\,\text{V}

(b) Similarly,

I=im2I = \frac{i_m}{\sqrt{2}}

So,

im=2 I=2×10≈14.14 Ai_m = \sqrt{2}\,I = \sqrt{2}\times 10 \approx 14.14\,\text{A}

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7.3A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.Show solution
For a pure inductor,

XL=ωL=2πνLX_L = \omega L = 2\pi \nu L

Given L=44 mH=44×10−3 HL = 44\,\text{mH} = 44\times 10^{-3}\,\text{H} and ν=50 Hz\nu = 50\,\text{Hz}:

XL=2π(50)(44×10−3)X_L = 2\pi(50)(44\times 10^{-3})

XL≈13.82 ΩX_L \approx 13.82\,\Omega

The rms current is

I=VXL=22013.82≈15.9 AI = \frac{V}{X_L} = \frac{220}{13.82} \approx 15.9\,\text{A}

So the rms current is **15.9 A15.9\,\text{A}**.

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7.4A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.Show solution
For a pure capacitor,

XC=1ωC=12πνCX_C = \frac{1}{\omega C} = \frac{1}{2\pi \nu C}

Given C=60 μF=60×10−6 FC = 60\,\mu\text{F} = 60\times 10^{-6}\,\text{F} and ν=60 Hz\nu = 60\,\text{Hz}:

XC=12π(60)(60×10−6)X_C = \frac{1}{2\pi(60)(60\times 10^{-6})}

XC≈44.21 ΩX_C \approx 44.21\,\Omega

Then

I=VXC=11044.21≈2.49 AI = \frac{V}{X_C} = \frac{110}{44.21} \approx 2.49\,\text{A}

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7.5In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
7.6A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
7.7A 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 the circuit, what is the average power transferred to the circuit in one complete cycle?
7.8Figure 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.

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Frequently Asked Questions

What are the important topics in Alternating Current for CBSE Class 12 Physics?
Alternating Current covers several key topics that are frequently asked in CBSE Class 12 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Alternating Current — CBSE Class 12 Physics?
Understand the core concepts first, then work through the 110 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Alternating Current Class 12 Physics?
This page has free step-by-step NCERT Solutions for every exercise question in Alternating Current (CBSE Class 12 Physics) — written the way examiners award marks: given, formula, working, answer.

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