Question

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

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

Answer 1

Answer: (A)

Current in L-C-R series circuit,
$i=\frac{V}{\sqrt{R^2+(X_L-X_C{)}^2}}$
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, $L=\frac{1}{(1000{)}^2\times 10\times 10^{-6}}$
= 0.1 H
= 100 m H.