Question

In $LCR$ circuit ${f = \frac{ 50 }{ \pi} \, Hz, V = 50 \, volt, R = 300 \, \Omega}$ . If $L = 1 \, H$ and $C = 20 \, \mu \, C$, then voltage across capacitor is

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

Answer 1

Answer: (D)

50 V

Answer 2

For an L-C-R circuit the impedance (Z) is given by

Z = $\sqrt{ R^2 + ( X_L - X_C)^2 }$
where $X_L = \omega L = 2 \pi f L$
and $X_C = \frac{ 1}{ \omega C} = \frac{ 1} { 2 \pi f C}$
Given, f = $\frac{ 50}{ \pi} \, Hz, R = 300 \Omega, \, L = 1 H$,
and C = 20 $\mu \, C = 20 \times 10^{ - 6 }$ C
$\therefore Z = \sqrt{ ( 300)^2 + \Bigg( 2 \pi \times \frac{ 50}{ \pi} \times 1 - \frac{ 1}{ 2 \pi \times \frac{ 50}{ \pi} \times 20 \times 10^{ - 6} } \Bigg) }$
Z = $\sqrt{ 90,000 + ( 100 - 500)^2 }$
Z = $\sqrt{ 90,000 + 16, 0000 } = 500 \Omega$
Hence, current in circuit is given by
i = $\frac{ V}{ Z} = \frac{ 50}{ 500}$
= 0.1 A
Voltage across capacitor is,
$V_C = i X_C = \frac{ 1}{ 2 \pi f C} = \frac{ 0.1}{ 2 \pi \times \frac{ 50}{ \pi} \times 20 \times 10^{ - 6} }$
= $\frac{ 0.1 \times 10^6 }{ 100 \times 20 }$
$\Rightarrow V_C = 50 \,V$