Question
Physics Question on Combination of capacitors
Three capacitors 3μF,6μF and 6μF are connected in series to a source of 120V. The potential difference, in volt, across the 3μF capacitor will be
A
24
B
30
C
40
D
60
Answer
60
Explanation
Solution
The combination of three charges in series C1=C11+C21+C31 C1=31+61+61 ⇒C=46=1.5μF The charge of this circuit q=CV=1.5×120 q=180μC The potential difference across the 3μF q=CV V=Cq=3180=60V