Question

What will be the capacitance of a capacitor in series order that can facilitate a lamp of 100 V, 50 W to connect to AC source of v = 200 sin 100t?
1. 10 μF
2. 50 μF
3. 100 μF
4. 5 μF

Answer 1

Correct Answer - Option 2 : 50 μF

Given: Rating of Lamp is 100 V, 50 W

P = V_R × I = 50 W

Where V_R is the Voltage rating of Lamp

The current flowing through the lamp will be the same as the current flowing through the circuit because the capacitor is used in series with the Lamp for facilitating the lamp

The current flow through the circuit is given by

I = 50/100 = 0.5 A

Now, V_R = 100 volt

\(V = \dfrac{200}{\sqrt2}\ volt\)

Consider Voltage across capacitor is V_c

For RC series Circuit, 

V_s² = V_R² + V_c²

\(V_c=\sqrt{V_s^2-V_R^2}\)

\(V_c=\sqrt{(\frac{200}{\sqrt2})^2-100^2}\)

V_c = 100 V

I × X_c = 100

X_c = 100/I = 100/0.5

= 200 Ω 

Therefore,

\(\dfrac{1}{ω C}=200\ \Omega\)

\(C = \dfrac{1}{200ω}\)

Given ω = 100 r/s

\(C = \dfrac{1}{200\times100}= 50\ \mu F\)