Question

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

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

Answer 1

Answer: (A)

In resonance condition, maximum current flows in the circuit.
Current in $LCR$ series circuit,
$i=\frac{V}{\sqrt{R^2+{\left(X_L-X_C\right)}^2}}$
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 ie,
$\omega L=\frac{1}{\omega C}$
or $L=\frac{1}{{\omega }^{2}C}\dots \left(i\right)$
Given, $\omega =1000s^{-1}$, $C=10\mu F=10\times 10^{-6}F$
Hence, $L=\frac{1}{{\left(1000\right)}^2\times 10\times 10^{-6}}$
$=0.1H$
$=100mH$