Question

In an $L–R$ series circuit $(L=\frac{175}{11}mHandR=12\Omega )$, a variable emf source $(V=V_{0}sin\omega t)$ of $V_{rms}=130\sqrt{2}V$ and frequency $50Hz$ is applied. The current amplitude in the circuit and phase of current with respect to voltage are respectively $(\text{use}\pi =\frac{22}{7})$.

• A. $20A,ta{n}^{-1}\left(\frac{5}{12}\right)$
• B. $10A,ta{n}^{-1}\left(\frac{5}{12}\right)$
• C. $10\sqrt{2}A,ta{n}^{-1}\left(\frac{5}{12}\right)$
• D. $10\sqrt{2}A,ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Answer 1

The correct option is A

$20A,ta{n}^{-1}\left(\frac{5}{12}\right)$

Given $L=\frac{175}{11}mHandR=12\Omega$
$\omega =2\pi f=100\pi$
Impedence in a $L-R$ Circuit$Z=\sqrt{(\omega L{)}^2+R^2}=13\Omega \text{Voltage amplitude}{V}_0=V_{rms}\times \sqrt{2}\Rightarrow V_0=130\times 2\text{Current amplitude}{I}_0=\frac{V_0}{Z}=\frac{130\times 2}{13}=20A\text{Phase angle}\varphi =ta{n}^{-1}\left(\frac{X_L}{R}\right)=ta{n}^{-1}\left(\frac{5}{12}\right)$