Question

For an LCR series circuit with an ac source of angular frequency $\omega$,

• A. Circuit will be capacitive if $\omega >\frac{1}{\sqrt{LC}}$
• B. Circuit will be capacitive if $\omega =\frac{1}{\sqrt{LC}}$
• C. Power factor of circuit will be unity if capacitive reactance equals inductive reactance
• D. Circuit will be leading voltage if $\omega >\frac{1}{\sqrt{LC}}$

Answer 1

The correct option is C Power factor of circuit will be unity if capacitive reactance equals inductive reactance

Considering option(a)
If $\omega >\frac{1}{\sqrt{LC}}$ then
${\omega }^2>\frac{1}{LC}$
$\Rightarrow \omega L>\frac{1}{\omega C}$
which tells us that circuit is inductive. So option(a) is incorrect.

Taking option(b)
If $\omega =\frac{1}{\sqrt{LC}}$
Then its the condition of resonance
$X_L=X_C$
And hence the circuit will behave as purely resistive. So option(b) is incorrect.

Taking option(c)
If $X_L=X_C$
then impedence $Z=R$
and we know that
$\cos \varphi =\frac{R}{Z}=\frac{R}{R}=1$
$\therefore$ option(c) is correct

Taking option(d)
We have already discussed in option(a) that in this condition circuit will behave as inductive circuit and in purely inductive circuit voltage leads the current. So option(d) is also incorrect.