Question

The value of inductance L for which the current is maximum in series LCR circuit with C = 10 $\mu$F and $\omega$ = 1000 rad/s

• A. 10 mH
• B. 50 mH
• C. 200 mH
• D. 100 mH

Answer 1

The correct option is D 100 mH

Maximum current flows in the circuit in resonance condition Current in the LCR circuit
$i=\frac{v}{\sqrt{R^2+(X_L-X_C{)}^2}}$
For current to be maximum denominator should be minimum
$(X_L-X_C{)}^2=0$
$\Rightarrow X_L=X_C\Rightarrow \omega L=\frac{1}{\omega C}$
$\therefore L=\frac{1}{{\omega }^{2}C}=\frac{1}{(100{)}^2\times 10\times {10}^{-6}}$
$L=\frac{1}{10}H=0.1H=100mH$