Question

A series circuit consists of a resistance of $15\Omega$, an inductance of $0.08\text{H}$ and a capacitor of capacitance $30\mu \text{F}$. The applied voltage has a frequency of $500\text{rad/s}$. Does the current lead or lag the applied voltage and by what angle. $\left(\text{given}{\tan }^{-1}\right(1.78)={60.6}^{\circ })$

• A. Lags by ${60.6}^{\circ }$
• B. Leads by ${60.6}^{\circ }$
• C. Lags by ${30.3}^{\circ }$
• D. Leads by ${30.3}^{\circ }$

Answer 1

The correct option is B Leads by ${60.6}^{\circ }$

The inductive reactance of the circuit is given as,

$X_L=\omega L=500\times 0.08=40\Omega$

The capacitive reactance of the circuit is given as,

$X_C=\frac{1}{\omega C}=\frac{1}{500\times (30\times {10}^{-6})}$

$X_C=66.7\Omega$


As capacitive reactance is more so, in circuit, current will lead the applied voltage by an angle $\varphi$ which is given as

$\tan \varphi =\left(\frac{\left(\frac{1}{\omega C}\right)-\omega L}{R}\right)$

$=\frac{66.7-40}{15}=1.78$

$\Rightarrow \varphi ={\tan }^{-1}1.78={60.6}^{\circ }$

Hence, option $\left(B\right)$ is correct.