A capacitance of $2 \mu \mathrm{~F}$ is required in an electrical circuit across a potential difference of 1.0 kV . A large number of $1 \mu \mathrm{~F}$ 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 :

JEE Main · 2017 · 02 Apr 2017 · Offline · Capacitance

A capacitance of

2 μ F

is required in an electrical circuit across a potential difference of 1.0 kV . A large number of

1 μ F

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 :

(A)24
(B)32✓ Correct
(C)2
(D)16

Answer: (B)

Solution

No single capacitor tolerates more than

300

V, so holding off

1000

V needs at least four in series, since

4 \times 300 = 1200

V. Four

1 μ

F units in series give

\frac{1}{4} = 0.25 μ

F per string. To rebuild

2 μ

F you then set

\frac{2}{0.25} = 8

such strings in parallel. That makes

4 \times 8 = 32

capacitors in all.

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option

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