Question 389: AC fundamentals

Question

The coil of an a.c. generator has 100 turns, each of cross-sectional area $$2 \mathrm{~m}^2$$. It is rotating at constant angular speed $$30 ~\mathrm{rad} / \mathrm{s}$$, in a uniform magnetic field of $$2 	imes 10^{-2} \mathrm{~T}$$. If the total resistance of the circuit is $$600 ~\Omega$$ then maximum power dissipated in the circuit is

Accepted Answer

$$\begin{aligned}
& N=100, A=2 \mathrm{~m}^2, \omega=30 \mathrm{rad} / \mathrm{s} \
& B=2 	imes 10^{-2} \mathrm{~T}, \mathrm{R}=600 \Omega
\end{aligned}$$
Maximum power dissipated in the circuit
$$\begin{aligned}
P_{\max }=E_{r m s} 	imes I_{r m s} & =\frac{E_0}{\sqrt{2}} 	imes \frac{I_0}{\sqrt{2}} \
& =\frac{E_0 I_0}{2} \quad 	ext{..... (i)}
\end{aligned}$$
But $$\mathrm{I}_0=\frac{\mathrm{E}_0}{\mathrm{R}}$$ ..... (ii)
Putting (2) into (1) we get,
$$\mathrm{I}_0=\frac{\mathrm{E}_0{ }^2}{2 \mathrm{R}}$$
But $$\mathrm{E}_0=\mathrm{NAB} \omega$$
$$\begin{aligned}
\mathrm{E}_0 & =100 	imes 2 	imes 2 	imes 10^{-2} 	imes 30 \
& =120 \mathrm{~V} \
	herefore \quad \mathrm{P}_{\max }= & \frac{120 	imes 120}{2 	imes 600}=12 \mathrm{~W}
\end{aligned}$$

Answer

$$6 \mathrm{~W}$$

Answer

$$9 \mathrm{~W}$$

Answer

$$12 \mathrm{~W}$$

Answer

$$24 \mathrm{~W}$$

AC fundamentals

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