Question

What is the value of inductance $L$ for which the current flowing is maximum in a series LCR circuit with $C=10\mu F$ and $\omega =1000s^{-1}$ ?

• A. $100mH$
• B. $1mH$
• C. Cannot be calculated unless $R$ is known
• D. $10mH$

Answer 1

The correct option is A $100mH$

Current in LCR series circuit,
$i=\frac{V}{\sqrt{R^2+{(X_L-X_C)}^2}}$
where $V$ is rms value of current, $R$ is resistance, $X_L$ is inductive reactance and $X_C$ is capacitive reactance.
For current to be maximum, denominator should be minimum which can be done, if
$X_L=X_C$
This happens in resonance state of the circuit i.e.,
$\omega L=\frac{1}{\omega C}$
or $L=\frac{1}{{\omega }^{2}C}$ .....(i)
Given, $\omega =1000s^{-1},C=10\mu F$
$=10\times {10}^{-6}F$
Hence, $L=\frac{1}{{\left(1000\right)}^2}\times 10\times {10}^{-6}$
$=0.1H$
$=100mH$