Question

In the circuit shown in the figure the EMF of battery is 100 V, resistance R is 50 $\Omega$ and capacitor C is 0.5 $\mu F$. Switch S is closed for long time and here is no voltage across capacitor. After switch is opened a maximum voltage of 100 V is found across capacitor then inductance L is

• A. 2.5 mH
• B. 1.25 mH
• C. 0.75 mH
• D. 0.25 mH

Answer 1

The correct option is B 1.25 mH


E=$i$R $\Rightarrow 100=i\times 50$

$i$=2A



By Conservation of internal energy $\Rightarrow \frac{1}{2}L{i}^2=\frac{1}{2}C{V}^2$

$L\times 2^2=0.5\times {10}^{-6}\times (100{)}^2\Rightarrow L=\frac{1}{8}\times {10}^{-2}=1.25mH$