Question

In the circuit shown in the figure, the AC source gives a voltage $V=10\sin \left(1000t\right)$. Neglecting source resistance, the voltmeter and ammeter reading will be

• A. $20\sqrt{\frac{125}{538}}V$ and $\frac{10}{\sqrt{538}}A$
• B. $10\sqrt{\frac{250}{528}}V$ and $\frac{10}{\sqrt{538}}A$
• C. $\sqrt{\frac{250}{538}}V$ and $\frac{1}{\sqrt{538}}A$
• D. $\sqrt{\frac{125}{539}}V$ and $\frac{1}{\sqrt{538}}A$

Answer 1

The correct option is B $10\sqrt{\frac{250}{528}}V$ and $\frac{10}{\sqrt{538}}A$

Impedance of the circuit is
$Z=\sqrt{R^2+(X_L-X_C{)}^2}$
$R=8\Omega$ (given)
$X_L=\omega L=1000\times 10\times {10}^{-3}=10\Omega$
$X_C=\frac{1}{\omega C}=\frac{1}{2000\times 20\times {10}^{-6}}=25\Omega$
$\Rightarrow Z=\sqrt{64+(15{)}^2}=\sqrt{64+225}=\sqrt{269}$
$I_{max}=\frac{V_0}{Z}=\frac{10}{\sqrt{260}}A$
$I_{rms}=\frac{I_{max}}{\sqrt{2}}=\frac{10}{\sqrt{538}}A$
$V_{rms}=I_{rms}\times \sqrt{5^2+(15{)}^2}$
$=\frac{10}{\sqrt{538}}\times \sqrt{25+225}=10\sqrt{\frac{250}{538}}V$.