Question

In L-C-R circuit, f=50/π Hz. V= 50V, R =300Ω. If L = 1H and C=20µC, then the voltage across the capacitor is :
A. 50V
B. 20V
 C. Zero
 D. 30V

Answer 1

Hint Try to calculate the impedance for the given circuit and then calculate the reactance of each capacitor and inductor and then proceed towards the question. Then find current flowing in the branch of the capacitor. As you are able to find these quantities you will be able to calculate the voltage across the given capacitor.

 Complete step-by-step solution:In the given question we are asked to calculate Voltage across the capacitor. For this we should know what is current flowing through that branch. To find current we will first calculate impedance for the given circuit
Formula for Impedance \[Z = \sqrt {{R^2} + {{({X_c} + {X_L})}^2}} \]
We know that, reactance of inductor and capacitor are given by
\[{X_L} = \omega L = 2\pi fL\]
\[{X_C} = \dfrac{1}{{\omega C}} = \dfrac{1}{{2\pi fC}}\]
\[f = \dfrac{{50}}{\pi }Hz,R = 300\Omega ,L = 1H\]
\[C = 20\mu C = 20 \times {10^{ - 6}}C\]
On substituting the values we get
\[Z = \sqrt {{{(300)}^2} + {{(2\pi \times \dfrac{{50}}{\pi } \times 1 - \dfrac{1}{{2\pi \times \dfrac{{50}}{\pi } \times 20 \times {{10}^{ - 6}}}})}^2}} \]
\[Z = \sqrt {90000 + {{(100 - 500)}^2}} \]
\[Z = \sqrt {90000 + 160000} \]
\[Z = \sqrt {250000} \]
On further solving we get
\[Z = 500\Omega \]
Hence the current in the circuit is given by
\[i = \dfrac{V}{Z} = \dfrac{{50}}{{500}} = 0.1A\]
Voltage across capacitor is
\[{V_c} = i{X_c} = \dfrac{i}{{2\pi fC}} = \dfrac{{0.1}}{{2\pi \times \dfrac{{50}}{\pi } \times 20 \times {{10}^{ - 4}}}}\]
\[{V_c} = \dfrac{{0.1 \times {{10}^6}}}{{100 \times 20}} = 50V\]

Hence, option A is correct.

Note: Reactance (X) is a measure of the opposition of capacitance and inductance to current. Reactance varies with the frequency of the electrical signal. Reactance is measured in ohms (ohm). There are two types of reactance: capacitive reactance and inductive reactance. The total reactance (X) is the difference between the two.