Question

For the AC circuit shown, the reading of ammeter and voltmeter are $5\text{A}$ and $50\sqrt{5}$ volts respectively, then

• A. average power delivered by the source is $250\text{W}$
• B. rms value of AC source is $50\text{volts}$
• C. rms value of AC source is $100\text{volts}$
• D. frequency of ac source is $\frac{100}{2\pi }\text{Hz}$

Answer 1

Answer: (A), (B)

The correct options are
A average power delivered by the source is $250\text{W}$
B rms value of AC source is $50\text{volts}$

Reading of ammeter $i_{\text{rms}}=5\text{A}$
Reading of voltmeter $=i_{\text{rms}}\sqrt{R^2+X_C^2}$
$50\sqrt{5}=5\sqrt{{10}^2+X_C^2}$
$500=100+X_C^2$
From this $X_C=20\Omega$
Since $X_C=\frac{1}{\omega C}=20\Omega$
$\omega =\frac{1}{X_{C}C}=\frac{1}{20\times 50\times {10}^{-6}}$
$\omega ={10}^3\text{rad/s}$

$\therefore X_L=\omega L={10}^3\times 20\times {10}^{-3}=20\Omega$
Therefore the circuit is in resonance,
We know that
$E_{\text{rms}}=i_{\text{rms}}R=50\text{V}$
Also power factor $=\cos \varphi =1$
Average Power $=E_{\text{rms}}{i}_{\text{rms}}\cos \varphi =250\text{W}$
Voltage gain $=\frac{X_L}{R}=2$