Question

Two circuits A and B are connected to identical d.c. sources each of e.m.f. 10 volt. Self-inductances are $L_A = 10$ H and $L_B = 10$ 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 ______.

• A. 800
• B. 1000
• C. 1200
• D. 1400

Hint:

If voltage and resistance are the same, the final current is the same. The energy required to "charge up" an inductor is purely dependent on its inductance size $L$!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the "ratio of energy consumed... to build up the current". In physics literature for LR circuits, the extra energy delivered by the battery specifically to establish the steady magnetic field against back-EMF is directly proportional to the energy ultimately stored in the inductor. We must compare this stored energy for two circuits.

Step 2: Detailed Explanation:
The steady-state maximum current ($I_0$) flowing through an LR circuit connected to a DC source depends only on the voltage and resistance, not the inductance.
$I_0 = \frac{V}{R}$
Since both circuits have identical voltage ($V = 10$ V) and identical resistance ($R = 40 \ \Omega$), they will ultimately reach the exact same steady-state current:
$I_{0A} = I_{0B} = \frac{10}{40} = 0.25$ A.
The energy ($U$) consumed from the source to build up the magnetic field and stored in the inductor is given by the formula:
$U = \frac{1}{2} L I_0^2$
Since $I_0$ is identical for both circuits, the energy consumed is directly proportional to the self-inductance:
$U \propto L$
Set up the ratio:
$\text{Ratio} = \frac{U_A}{U_B} = \frac{L_A}{L_B}$
Substitute the given inductances (making sure units match):
$L_A = 10$ H
$L_B = 10 \text{ mH} = 10 \times 10^{-3} \text{ H}$
$\text{Ratio} = \frac{10}{10 \times 10^{-3}}$
$\text{Ratio} = \frac{1}{10^{-3}} = 10^3 = 1000$

Step 3: Final Answer:
The ratio of energy consumed is 1000, matching option (b).