Question 302: AC fundamentals

Question

A light bulb connected in series with a capacitor and an a.c. source is glowing with a certain brightness. On reducing the capacity of capacitance and frequency of source, the brightness of the lamp (respectively)

Accepted Answer

When a light bulb is connected in series with a capacitor and an AC source, the current (and thus the bulbÃ¢ÂÂs brightness) depends on the
capacitive reactance
$X_C$, which is given by:
$ X_C \;=\; \frac{1}{\omega C} \;=\; \frac{1}{2\pi f \,C}. $
If $C$
decreases
, $X_C$
increases
, causing the current to
decrease
, so the bulbÃ¢ÂÂs brightness
reduces
.
If $f$ (frequency)
decreases
, $X_C$
increases
again (because $f$ is in the denominator), causing the current to
decrease
, so the bulbÃ¢ÂÂs brightness
reduces
.
Hence, on
reducing both the capacitance and the frequency
, the brightness
decreases
in each case.
The correct answer is:
Option B: is reduced, is reduced.

Answer

is reduced, is increased

Answer

is reduced, is reduced

Answer

is increased, is reduced

Answer

is increased, is increased

AC fundamentals

A light bulb connected in series with a capacitor and an a.c. source is glowing with a certain brightness. On reducing the capacity of capacitance and frequency of source, the brightness of the lamp (respectively)

Choose the correct answer: