Question

In a series $\text{RLC}$ circuit $R=300\Omega ,L=60\text{mH},C=0.5\mu \text{F},V=50\text{V},\omega ={10}^4\text{rad}{\text{s}}^{-1}$.Find the voltage across capacitor:

• A. 30 V
• B. 20 V
• C. 60 V
• D. 50 V

Answer 1

Answer: (B)

20 V

Current in the circuit is given by:

$I=\frac{V}{|Z|}$

$I=\frac{V}{\sqrt{R^2+(\omega L-\frac{1}{\omega C}{)}^2}}$

$I=\frac{50}{\sqrt{{300}^2+{(60\times {10}^{-3}\times {10}^4-\frac{1}{0.5\times {10}^4\times {10}^{-6}})}^2}}$

$I=\frac{50}{\sqrt{{300}^2+{400}^2}}=\frac{1}{10}$A

Voltage across capacitor is:

$V_c=I{X}_c=\frac{I}{\omega C}=\frac{1}{10}\times 200=20V$