Question

A tungsten filament bulb rated at 500w, 100v is to be connected in series to a capacitor across 220v, 50hz supply. Calculate the capacitor such that the voltage and power consumed by the bulb are according to the rating of the bulb

Answer 1

Explanations & Calculations

• This symbolizes an RC circuit.

• The current flowing in the circuit can be found by the bulb assuming it is a resistor,

$\qquad\qquad \begin{aligned} \small I&=\small \frac{P}{V}=\frac{500W}{100V}=5A \end{aligned}$

• If the voltage across the cap is $\small V_c$, then the phasor sum of the voltages will be

$\qquad\qquad \begin{aligned} \small \vec{V_b}+\vec{V_c}&=\small 220V \end{aligned}$

• Then since they are orthogonal to each other,

$\qquad\qquad \begin{aligned} \small \sqrt{V_b^2 +V_c^2 }&=\small 220\\ \small V_c&=\small \sqrt{220^2 -100^2}=195.96V \end{aligned}$

• Then, by the equation for a capacitor

$\qquad\qquad \begin{aligned} \small V_c &=\small IX_c=\frac{I}{2\pi f C}\\ \small C&=\small \frac{I}{2\pi f V_c}\\ &=\small \frac{5A}{2\pi\times 50Hz\times 195.96V}\\ &=\small \bold{81.22\,\mu F} \end{aligned}$