Question

In the figure shows $R=100\Omega ,L=\frac{2}{\pi }\text{H}$, and $C=\frac{8}{\pi }\mu \text{F}$ are connected in series with an a. c. source of $200\text{volts}$ and frequency $f$. $V_1$ and $V_2$ are two hot-wire voltmeters. If the readings of $V_1$ and $V_2$ are same then which of the following is not correct.

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

Answer 1

The correct option is D $f=250\pi \text{Hz}$

Given,
$R=100\Omega$

$L=\frac{2}{\pi }\text{H}$

$C=\frac{8}{\pi }\mu \text{F}$

According to question , $V_1=V_2\text{Or}{V}_L=V_C$

$\Rightarrow X_L=X_C\Rightarrow \omega L=\frac{1}{\omega C}$

${\omega }^2=\frac{1}{LC}=\frac{1}{\frac{2}{\pi }\times \frac{8}{\pi }\times {10}^{-6}}$

${\omega }^2=(\frac{\pi }{4}\times {10}^3{)}^2\Rightarrow \omega =\frac{1000\pi }{4}\text{or}250\pi$

$\therefore 2\pi f=250\pi \text{and}f=125\text{Hz}$

Option (a) is correct.

Now current through the R is
$I_{rms}=\frac{V_{rms}}{Z}$= $\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C{)}^2}}$

$I_{rms}=\frac{Vrms}{R}$ $\because X_L=X_C$

$I_{rms}=\frac{200}{100}=2\text{A}$

And as $V_1=V_2$

$\Rightarrow V_C=V_L=\omega L=2\pi fL=2\pi \times 125\times 2=1000\text{volt}$

Hence, option (b) is the correct alternative.