Question

An inductor coil wound uniformly has self inductance 'L' and resistance 'R'. The coil is broken into two identical parts. The two parts are then connected in parallel across a battery of 'E' volt of negligible internal resistance. The current through battery at steady state is

• A. $\frac{2E}{R}$
• B. $\frac{3E}{R}$
• C. $\frac{4E}{R}$
• D. $\frac{E}{R}$

Hint:

Cutting a wire in half divides its individual resistance by 2. Connecting two identical resistors in parallel cuts the resistance in half once more! This drops the total network resistance to one-quarter ($\frac{R}{4}$), which automatically quadruples the total steady-state circuit current to $\frac{4E}{R}$.

Answer 1

The Correct Option is C

At a steady state under a constant D.C. voltage source, an inductor acts as a simple ideal conducting wire since the rate of current change is zero ($\frac{di}{dt} = 0$), making the inductive reactance vanish completely ($X_L = 0$). Therefore, only the purely ohmic resistance properties dictate the circuit current. 1. When the uniform coil is cut into two identical halves, the resistance of each individual segment scales proportionally to half of the initial total: $$R_1 = R_2 = \frac{R}{2}$$ 2. Connecting these two identical half-coils in a parallel branch configuration updates the net equivalent circuit resistance ($R_p$) via: $$\frac{1}{R_p} = \frac{1}{\left(\frac{R}{2}\right)} + \frac{1}{\left(\frac{R}{2}\right)} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R} \implies R_p = \frac{R}{4}$$ Applying Ohm's law to evaluate the steady-state current ($I$) drawn from the battery source: $$I = \frac{E}{R_p} = \frac{E}{\left(\frac{R}{4}\right)} = \frac{4E}{R}$$
Final Answer:
The current through the battery at steady state is $\frac{4E}{R}$, which corresponds to option (C).