Question

In the figure shown the battery is ideal. The values are $\epsilon =10V,R=5\Omega,L=2H$ . The current through the battery at $t=2s$ after closing the switch is:

• A. $12A$
• B. $7A$
• C. $3A$
• D. None of these

Answer 1

Answer: (A)

From inductor, potential difference $V_{L}=L\frac{d i_{L}}{d t}\Rightarrow 10=2\frac{d i_{L}}{d t}$
Current through inductor, $i_{L}=5t$
At $t=2s$ , $i_{L}=10A$
Current through resistance, $i_{R}=2A$
Current through battery, $i=i_{L}+i_{R}=12A$