Question

A $50W,{\text{ }}100V\;$lamp is to be connected to an AC mains of $200V,{\text{ }}50Hz$. What capacitance is essential to be put in series with the lamp?
A. $9.2$
B. $8$
C. $10$
D. $12$

Answer 1

Initially, if we consider combining voltages to obtain the out of phase voltage across the capacitor. Then to apply the relationship between the current and the AC voltage across the capacitor of a given frequency. We find the unknown capacitance value that is connected in series.

Complete step by step solution:
The voltage necessary across the lamp is $100V$ and at this voltage, the current through both bulb and capacitor will be $I = 0.5A$ to dissipate $50W$ in the bulb. First we need the voltages through the bulb and capacitor to combine to give$200V$. Along with the current through each being in phase then the voltage $V$ across the capacitor will be $90^\circ$ out of phase with that across the bulb which is incandescent and therefore entirely resistive. So we want to use the Pythagoras theorem to combine them both:
${100^2} + {V^2} = {200^2}$And so $V = 100\sqrt 3$
Now the equation for the relationship between AC voltage and current for a capacitor is
$I = 2\pi fCV$ Where frequency $f = {\text{ }}50Hz$ in this case. Calculating for $C$ and putting the values for $I\;$and $V$ estimated above we get:
$C = \dfrac{I}{{2\pi fV}}$
$= \dfrac{{0.5}}{{1000\pi \sqrt 3 }}$
$= 50\pi \sqrt 3$
This is about $9.19\mu F$ which is the first option A.

Note:
It is more economical to apply inductance and capacitance in series with the lamp for it to function. This is because it consumes no power as there can be a dissipation of power when resistance is introduced in series to the lamp. As the capacitance increases, the current also increases and the bulbs become brighter. If the frequency of the AC power source increases, the current and the brightness also increases. On a DC source, the current will flow only for a restricted short time. The higher the impedance of the bulb or a lamp, the lower its current, and the higher the efficiency of the bulb becomes the higher its brightness.