Question

A series $RLC$ circuit is driven by a generator at frequency $1000\text{Hz}$. The inductance is $90.0\text{mH},$ Capacitance is $0.500\mu \text{F}$ and the phase constant has a magnitude of ${60}^{\circ }$ (Take ${\pi }^2=10$)

• A. Here current leads the voltage
• B. Here voltage leads the current
• C. Resistance of circuit is $\frac{80\pi }{\sqrt{3}}\Omega$
• D. At resonance $X_L=\omega L=180\pi \Omega$

Answer 1

Answer: (B), (C), (D)

At resonance $X_L=\omega L=180\pi \Omega$

Given
$f=1000\text{Hz}$
$L=90\text{mH}$
$C=0.500\mu \text{F}$
$\varphi ={60}^{\circ }$
$X_L=\omega L=90\times {10}^3\times 2\pi \omega =180\pi \Omega$
$X_C=\frac{1}{\omega C}=\frac{1}{0.5\times {10}^{-6}\times 2\pi \times 1000}=\frac{1000}{\pi }\Omega$
The circuit is inductive as $X_L>X_C$. Then voltage leads the current.

Phase angle can be written as
$\tan \varphi =\frac{\omega L-\frac{1}{\omega C}}{R}=\frac{80\pi }{R}$
$\sqrt{3}=\frac{80\pi }{R}$
$R=\frac{80\pi }{\sqrt{3}}\Omega$