Question

In an ac circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of inductance is $0.5$ henry and of capacitance is $8\mu F$. The angular frequency of the input AC Voltage must be equal to

• A. $500$
• B. $5\times {10}^4$
• C. $4000$
• D. $5000$

Answer 1

The correct option is A $500$

Given,

$L=0.5H$

$C=8\mu F$

We know,

$I=\frac{V}{Z}$

$I=\frac{V}{\sqrt{R^2+(V_L-V_C{)}^2}}$

For I to be maximum, Z should be minimum,

$|V_L-V_C|$ should be minimum (as R can't be altered here).

But, $|V_L-V_C|>0$

The minimum value of above expression can be zero,

For that,

$V_L=V_C$

or, $\omega L=\frac{1}{\omega C}$

or, ${\omega }^2=\frac{1}{LC}$

or, $\omega =\frac{1}{\sqrt{LC}}$

or, $\omega =\frac{1}{\sqrt{0.5\times 8\times {10}^{-6}}}$

or, $\omega =\frac{1}{\sqrt{4\times {10}^{-6}}}$

or, $\omega =\frac{1}{2\times {10}^{-3}}=500s^{-1}$

Hence Option $A$ is correct.