Question

A $100\text{volt}$ AC source of frequency $500\text{Hz}$ is connected to a $\text{L–C-R}$ circuit with $L=8.1\text{mH}$, $C=12.5\text{µF}$ and $R=10\Omega$ , all connected in series. The potential difference across the resistance is -

• A. $400\text{V}$
• B. $300\text{V}$
• C. $200\text{V}$
• D. $100\text{V}$

Answer 1

The correct option is D $100\text{V}$

Given: $V=100\text{volt},f=500\text{Hz},L=8.1\text{mH},C=12.5\text{µF},R=10\Omega$

The impedance of L-C circuit is,

$Z=\sqrt{R^2+(X_L-X_c{)}^2}$

Here $X_L=2\pi fl=2\times 3.14\times 500\times (8.1\times {10}^{-3})=25.4\Omega$

and $X_c=\frac{1}{2\pi fC}=\frac{1}{2\times 3.14\times 500\times 12.5\times {10}^{-6}}=25.4\Omega$

$R=10\Omega$

$\therefore Z=\sqrt{(10{)}^2+(25.4-25.4{)}^2}=10\Omega$

Now, $i_{rms}=\frac{E_{rms}}{Z}=\frac{100}{10}=10A$

$\therefore V_R=i_{rms}\times R$

$V_R=10\times 10=100\text{V}$

Hence, option $A$ is correct.