Question
A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :
-
24
-
32
-
2
-
16
Solution
B.
32
To hold 1 KV potential difference minimum four capacitors are required in series
⇒ C1 = 1/4
for one series.
So for Ceq to be 2μF, 8 parallel combinations are required.
⇒ Minimum no. of capacitors = 8 × 4 = 32
