Question

In the given figure, a series $L-C-R$ circuits is connected to a variable frequency source of $230V$. The impedance and amplitude of the current at the resonating frequency will be

• A. $20\Omega$ and $4.2A$
• B. $30\Omega$ and $6.9A$
• C. $25\Omega$ and $5.8A$
• D. $40\Omega$ and $5.75A$

Answer 1

The correct option is D $40\Omega$ and $5.75A$

Inductance $L=5.0H$

Capacitance $C=80\times {10}^{-6}F$

Resistance $R=40\Omega$

Effective voltage $E_v=230V$

Now, the resonance angular frequency

${\omega }_r=\frac{1}{\sqrt{LC}}$

${\omega }_r=\frac{1}{\sqrt{5\times 80\times {10}^{-6}}}$

${\omega }_r=\frac{1}{\sqrt{400\times {10}^{-6}}}$

${\omega }_r=\frac{1}{20\times {10}^{-3}}$

${\omega }_r=50rad/s$

Now, impedance of the circuit

$Z=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$

At resonance

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

So,

$Z=\sqrt{R^2}$

$Z=R=40\Omega$

Now, amplitude of current at resonating frequency

Peak value of current

$I_0=\frac{E_0}{Z}$

$I_0=\frac{\sqrt{2}\times E_v}{Z}$

$I_0=\frac{\sqrt{2}\times 230}{40}$

$I_0=8.13A$

Now, Rms value of current

$I_v=\frac{I_0}{\sqrt{2}}$

$I_v=\frac{8.13}{\sqrt{2}}$

$I_v=5.75A$

Hence, the impedance and amplitude of the current at the resonating frequency will be $40\Omega$ and $5.75A$