Question:

The coil of an a.c. generator has 100 turns, each of cross-sectional area $2 \text{ m}^2$. It is rotating at constant angular speed $30 \text{ rad/s}$, in a uniform magnetic field of $2\times10^{-2}\text{ T}$. If the total resistance of the circuit is 600 $\Omega$ then maximum power dissipated in the circuit is

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Logic Tip: You can bypass calculating the current by directly using the formula $P_{avg} = \frac{E_0^2}{2R}$. Substituting the values directly gives $P = \frac{(120)^2}{2 \times 600} = \frac{14400}{1200} = 12 \text{ W}$.
Updated On: Apr 28, 2026
  • 6 W
  • 9 W
  • 12 W
  • 24 W
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The Correct Option is C

Solution and Explanation

Concept:
As the coil rotates in a magnetic field, it induces an alternating e.m.f. given by Faraday's law. The peak induced e.m.f. ($E_0$) is $E_0 = NAB\omega$. The question asks for the average power dissipated (often termed simply as "power dissipated" in standard AC circuit contexts). The average power $P$ for a purely resistive circuit is $P = E_{rms} \times I_{rms} = \frac{E_0}{\sqrt{2 \times \frac{I_0}{\sqrt{2 = \frac{E_0^2}{2R}$.
Step 1: Calculate the peak induced e.m.f. ($E_0$).
Given data: Number of turns, $N = 100$ Area, $A = 2 \text{ m}^2$ Magnetic field, $B = 2 \times 10^{-2} \text{ T}$ Angular speed, $\omega = 30 \text{ rad/s}$ Use the formula for peak e.m.f.: $$E_0 = NAB\omega$$ $$E_0 = 100 \times 2 \times (2 \times 10^{-2}) \times 30$$ $$E_0 = 100 \times 10^{-2} \times 2 \times 2 \times 30$$ $$E_0 = 1 \times 4 \times 30 = 120 \text{ V}$$
Step 2: Calculate the average power dissipated.
Given the circuit resistance $R = 600 \text{ }\Omega$. The peak current is $I_0 = \frac{E_0}{R} = \frac{120}{600} = 0.2 \text{ A}$. The average power dissipated over a full cycle in an AC generator circuit is: $$P = \frac{E_0 I_0}{2}$$ $$P = \frac{120 \times 0.2}{2}$$ $$P = \frac{24}{2} = 12 \text{ W}$$ (Note: Standard competitive exam phrasing sometimes uses "maximum power dissipated" to refer to the average power capacity of the generator, distinguished from instantaneous peak power $E_0^2/R = 24 \text{ W}$).
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