Question

A lamp rated 500 W, 100 V is connected in series with a capacitor across 250 V, 50 Hz supply. Determine the value of the capacitance.​

Answer 1

The voltage required across the lamp is 100V and at this voltage the current through both bulb and capacitor will be I=0.5 A to dissipate 50W in the bulb.

First we need the voltages across bulb and capacitor to combine to give 200V. With the current through each being in phase then the voltage V across the capacitor will be 90° out of phase with that across the bulb (which I’m assuming is incandescent and therefore purely resistive). So we need to use Pythagoras to combine them:

1002+V2=2002

and so V=1003–√ .

Now the relationship between AC voltage and current for a capacitor is

I=2πfCV

where frequency f is 50Hz in this case. Solving for C and inserting the values for I and V calculated above we get:

C=I2πfV

=0.510000π3–√

=50π3–√μF

This is approximately 9.19μ F .

Answer 2

The rated values for bulb are :  

voltage = 100 V and

current I = $\frac{W}{V} = \frac{500}{100} =5A$.

Obviously, the bulb has been treated as a pure resistance : Vₐ =

 $\sqrt{250^{2} - 100^{2} } = \sqrt{62500-10000} \\\\\sqrt{62500-10000} = \sqrt{52500} \\\\\sqrt{52500} = 229.13V$

IXₐ = 229.13V

5.Xₐ = 229.13V

Xₐ = 45.826V

C = $\frac{1}{314} * 45.826$

C = 9.19μF