Minimum Number of Capacitors – Rankers Physics

Parallel Plate Capacitor: Practice Problem & Solution

From a supply of identical capacitors rated \( 8\ \mu\text{F}\ \), \( 250\text{V} \), the minimum number required to form a composite \( 16\ \mu\text{F}\ \), \( 1000\text{V} \) capacitor is
\( 2 \)
\( 4 \)
\( 8 \)
\( 32 \)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:

To handle \( 1000\text{V} \), \( N_s = \frac{1000}{250} = 4 \) capacitors must be connected in series. The capacitance of one row is \( C_s = \frac{8}{4} = 2\ \mu\text{F} \). To get a total capacitance of \( 16\ \mu\text{F} \), we need \( N_p = \frac{16}{2} = 8 \) parallel rows. Total number \( = N_s \times N_p = 4 \times 8 = 32 \).

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