Question 202: Electromagnetic Induction

Question

Two circuits A and B are connected to identical d.c. sources each of e.m.f. 10 volt. Self-inductances of circuits A and B are respectively $\mathrm{L}_{\mathrm{A}}=10 \mathrm{H}$ and $\mathrm{L}_{\mathrm{B}}=10 \mathrm{mH}$. The total resistance of each circuit is $40 \Omega$. The ratio of energy consumed in circuit A and circuit B to build up the current to steady value is

Accepted Answer

Both circuits use the same battery (10 V) and have the same resistance (40 ÃÂ©). This means the steady current ($I$) will be the same in both circuits.
The energy stored in the inductor in each circuit is given by:
$$ \mathrm{E} = \frac{1}{2} \mathrm{L} I^2 $$
To find the ratio of energy stored in the inductors of circuits A and B, we can write:
$$ \frac{E_A}{E_B} = \frac{L_A I^2}{L_B I^2} $$
The current $I$ is the same in both, so $I^2$ cancels out:
$$ \frac{E_A}{E_B} = \frac{L_A}{L_B} $$
Given: $L_A = 10 	ext{ H}$ and $L_B = 10 	ext{ mH} = 10 	imes 10^{-3} 	ext{ H}$
So,
$$ \frac{E_A}{E_B} = \frac{10}{10 	imes 10^{-3}} = 1000 $$

Answer

800

Answer

1000

Answer

1200

Answer

1400

Electromagnetic Induction

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