Question

In the circuit shown, resistance $R=100\Omega$, indutance $L=\frac{2}{\pi }$ Henry and capacitance $C=\frac{8}{\pi }\mu F$ are connected in series with an ac source of $200$ volt and frequency ${}^{\prime }{f}^{\prime }$. If the readings of the hot wire voltmeters $V_1$ and $V_2$ are same, then

• A. $V_1=V_2=1000$ Volt
• B. $f=125\text{Hz}$
• C. current through $R$ is $2\text{A}$
• D. $f=250\text{Hz}$

Answer 1

Answer: (A), (B), (C)

current through $R$ is $2\text{A}$

Given
$V_1=V_2\Rightarrow V_L=V_C$
$I{X}_L=I{X}_C$
$X_L=X_C$
$\omega L=\frac{1}{\omega C}$
$f=\frac{1}{2\pi \sqrt{LC}}$

Here
$R=100\Omega ,L=\frac{2}{\pi }\text{H}$
$C=\frac{8}{\pi }\mu F$ and $V=100$ Volts
$f=\frac{1}{2\pi \sqrt{\frac{2}{\pi }\times \frac{8}{\pi }\times {10}^{-6}}}=\frac{1}{2\sqrt{16\times {10}^{-6}}}$
$f=\frac{1}{8}\times {10}^3=125\text{Hz}$
$I_{rms}=\frac{V_{rms}}{Z}\text{here}Z=\sqrt{R^2+(X_L-X_C{)}^2}$
$Z=R$
$I_{rms}=\frac{200}{100}=2\text{A}$
$V_1=V_2=V_L=V_C$
$V_L=I_{rms}{X}_L=2\omega L$
$V_L=2\times 2\pi fL=2\times 2\pi \times 125\times \frac{2}{\pi }$
$V_L=8\times 125=1000\text{V}$
So $V_1=V_2=1000\text{V}$