Physics Class 12 Part I…
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Ac Voltage Resistor

🎓 Class 12 Physics CBSE Theory Ch 7 – Alternating Current ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Ac Voltage Resistor

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Ac Voltage Resistor

7.1 Introduction

Almost every electrical socket in our homes, schools and factories delivers alternating current (AC), not direct current. The reason is simple: AC voltage can be stepped up or down through a transformer, making long-distance power transmission cheap and efficient. In this chapter we study how AC voltage drives current through resistors, inductors and capacitors, the concept of impedance and resonance, the meaning of power factor, and finally the working of transformers.

7.2 AC Voltage Applied to a Resistor

Consider a source whose terminal voltage varies sinusoidally with time:

\(v = v_m \sin\omega t\)

Here \(v_m\) is the peak voltage (amplitude), \(\omega = 2\pi f\) is the angular frequency, and \(f\) is the frequency in hertz. This source is connected across a pure resistor of resistance \(R\) (Fig. 7.1).

v = vm sin ωt R i
Fig. 7.1: A sinusoidal AC source connected to a pure resistor R.

Applying Kirchhoff's loop rule around the circuit:

\(v = iR \;\Rightarrow\; i = \dfrac{v_m}{R}\sin\omega t = i_m \sin\omega t\)

where \(i_m = v_m/R\) is the peak current. Two key conclusions:

  • The current is also sinusoidal, with the same frequency as the voltage.
  • Current and voltage are in phase — both reach their maxima, zeros and minima at the same instants.
t v = vm sin ωt i = im sin ωt v, i
Fig. 7.2: In a pure resistor the voltage v (solid blue) and current i (dashed red) are perfectly in phase.

7.2.1 Power Dissipated and the Need for Averaging

The instantaneous power dissipated as heat in R is:

\(p = i^2 R = i_m^2 R \sin^2\omega t\)

Since \(\sin^2\omega t\) keeps changing, so does \(p\). To find the average power over one cycle, we use \(\langle \sin^2\omega t \rangle = 1/2\):

\(P = \langle p \rangle = \dfrac{1}{2}\,i_m^2 R\)

7.2.2 Root-Mean-Square (RMS) Values

We can rewrite the average power as \(P = (i_m/\sqrt{2})^2 R\). This motivates a new quantity, the root-mean-square (RMS) value:

\(I = i_{rms} = \dfrac{i_m}{\sqrt{2}} = 0.707\, i_m,\qquad V = v_{rms} = \dfrac{v_m}{\sqrt{2}} = 0.707\, v_m\)

In terms of these:

\(P = I^2 R = V I = \dfrac{V^2}{R}\)

This is exactly the form for DC power - which is why RMS values were invented. Whenever an AC ammeter or voltmeter reads "1 A" or "230 V" it is showing the RMS value.

Indian mains supply: The household AC voltage is \(V = 220\) V (rms) at \(f = 50\) Hz. The peak voltage is \(v_m = \sqrt{2} \times 220 \approx 311\) V and the period is \(T = 1/50 = 20\) ms.

7.2.3 Phasor Diagram

Sinusoidal quantities of the same frequency can be represented by rotating arrows called phasors. A phasor of length \(v_m\) (or \(i_m\)) rotates anticlockwise at angular speed \(\omega\); its vertical projection gives the instantaneous value. For a pure resistor the V-phasor and I-phasor lie along the same direction, confirming they are in phase.

V I O ω →
Fig. 7.3: V and I phasors for a pure resistor lie along the same line - phase difference = 0.
QuantitySymbolRelationIndian mains value
Peak voltage\(v_m\)= √2 V311 V
RMS voltageV= vm/√2220 V
Peak current\(i_m\)= vm/Rdepends on R
RMS currentI= im/√2depends on R
Frequencyfω/2π50 Hz

7.2.4 Worked Examples

Example 7.1 — Bulb on AC mains

A 100 W, 220 V bulb is connected to an AC source of peak voltage 311 V at 50 Hz. Find (a) the RMS current and (b) the peak current.

(a) Power \(P = VI\). Since \(P = 100\) W and \(V = 220\) V:

\(I = P/V = 100/220 = 0.455\) A (RMS)

(b) Peak current \(i_m = \sqrt{2}\,I = 1.414 \times 0.455 = 0.643\) A.

Example 7.2 — Heater element

The resistance of an electric heater is 100 Ω. Calculate (a) the RMS current drawn from a 220 V (rms), 50 Hz mains and (b) the average power dissipated.

(a) \(I = V/R = 220/100 = 2.20\) A.

(b) \(P = I^2 R = (2.20)^2 \times 100 = 484\) W.

The peak current is \(i_m = \sqrt{2} \times 2.20 = 3.11\) A and the instantaneous power oscillates between 0 and \(i_m^2 R = 968\) W — but the time average is 484 W.

Simulation: RMS Value Calculator

Drag the slider to set the peak voltage and resistance. The simulator computes \(v_{rms}\), \(i_m\), \(i_{rms}\) and average power dissipated.

RMS voltage V219.9 V
Peak current im3.11 A
RMS current I2.20 A
Average power P483.4 W
Activity 7.1 — Why a torch bulb glows the same on AC and DC

Set up: a 6 V torch bulb, a 6 V dry battery (DC), and a 6 V (RMS) low-voltage AC source.

Predict: will the bulb glow brighter on DC 6 V, on AC 6 V (RMS), or the same?
  1. Connect the bulb to the DC source. Note the brightness.
  2. Replace the DC with an AC source of 6 V (RMS).
  3. Compare.

The bulb glows with the same brightness in both cases. The 6 V RMS is precisely the steady DC value that delivers the same average power - that is the operational definition of RMS.

Competency-Based Questions L1L2L3L4L5

A laboratory AC source delivers a sinusoidal voltage of peak value 170 V at 50 Hz to a pure resistor of 85 Ω. The connecting wires are ideal.

1. The RMS voltage supplied is approximately: L1

  • (a) 85 V
  • (b) 120 V
  • (c) 170 V
  • (d) 240 V
(b) 120 V. \(V = v_m/\sqrt{2} = 170/1.414 \approx 120\) V.

2. Explain why voltage and current are in phase across a pure resistor. L2

Ohm's law \(i = v/R\) is instantaneous. Whenever v reaches its maximum, i = v/R also reaches its maximum. The peaks, zeros and minima of v and i coincide in time, so their phase difference is zero.

3. Calculate the average power dissipated in the resistor described in the scenario. L3

\(I = V/R = 120/85 = 1.41\) A. \(P = I^2 R = (1.41)^2 \times 85 \approx 169\) W. Alternatively, \(P = \tfrac{1}{2} i_m^2 R = \tfrac{1}{2}(170/85)^2 \times 85 = \tfrac{1}{2}\times 4 \times 85 = 170\) W (rounding).

4. An AC ammeter and a DC ammeter (moving-coil) are connected in series with an AC source. The AC ammeter reads 5 A. What will the DC ammeter read, and why? L4

The moving-coil DC ammeter responds to the average current. Over one cycle of a pure sinusoid the average is zero, so the DC ammeter reads zero. The AC ammeter (hot-wire or moving-iron type) responds to i², giving the RMS value 5 A.

5. A student claims that doubling the frequency of the AC source (with the same RMS voltage) doubles the power dissipated in a resistor. Evaluate this claim. L5

The claim is FALSE. For a pure resistor, average power \(P = V^2/R\) depends only on V (RMS) and R - it is independent of frequency. Frequency only affects circuits containing L or C.

Assertion-Reason Questions

Options: (A) Both A and R are true; R is the correct explanation of A. (B) Both true but R is not the correct explanation. (C) A true, R false. (D) A false, R true.

Assertion (A): The RMS value of an alternating current is always less than its peak value.

Reason (R): For a sinusoidal current, \(I_{rms} = i_m/\sqrt{2} \approx 0.707\, i_m\).

(A). Both true and R correctly explains A.

Assertion (A): An AC voltmeter and AC ammeter measure peak values directly.

Reason (R): Calibration is performed for sinusoidal RMS values.

(D). A is false (instruments are calibrated to display RMS, not peak). R is true.

Assertion (A): When a sinusoidal voltage is applied to a resistor, the average value of current over one full cycle is zero.

Reason (R): Positive and negative half-cycles of sinusoidal current have equal area, so they cancel on averaging.

(A). Both true and R correctly explains A. (RMS uses the mean of i², not of i.)

Frequently Asked Questions - Ac Voltage Resistor

What is the main concept covered in Ac Voltage Resistor?
In NCERT Class 12 Physics Chapter 7 (Alternating Current), "Ac Voltage Resistor" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Ac Voltage Resistor useful in real-life applications?
Real-life applications of "Ac Voltage Resistor" from NCERT Class 12 Physics Chapter 7 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Ac Voltage Resistor?
Key formulas in "Ac Voltage Resistor" (NCERT Class 12 Physics Chapter 7 Alternating Current) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 7?
NCERT Class 12 Physics Chapter 7 (Alternating Current) is structured so each part builds on the previous one. "Ac Voltage Resistor" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Ac Voltage Resistor?
CBSE board questions from "Ac Voltage Resistor" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Ac Voltage Resistor" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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Physics Class 12 Part I – NCERT (2025-26)
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