Question

In the figure shown, a circuit contains two identical resistors with resistance \(R=5~\Omega \) and an inductance with \(L=2~\text{mH}\). An ideal battery of \(15~\text V\) is connected in the circuit. What will be the current through the battery long after the switch is closed?


2R

• A. $3\text{A}$
• B. $5.5\text{A}$
• C. $6\text{A}$
• D. $7.5\text{A}$

Answer 1

The correct option is C $6\text{A}$

Long time after switch is closed, the inductor will be idle, so, the equivalent diagram will be as below


Since the resistors are parallel to each other, using Ohm's law we have,

$I=\frac{\epsilon }{\left(\frac{R\times R}{R+R}\right)}=\frac{2\epsilon }{R}=\frac{2\times 15}{5}=6\text{A}$