Question

An electrician requires a capacitance of 6 μF in a circuit across a potential difference of 1.5 kV. A large number of 2 μF capacitors which can withstand a potential difference of not more than 500 V are available. The minimum number of capacitors required for the purpose is

• A. 3
• B. 9
• C. 6
• D. 27

Answer 1

The Correct Option is D

1. Number of capacitors in series:

To withstand the $1500\,\text{V}$ potential difference, we need to connect capacitors in series. The voltage across each capacitor in a series connection adds up to the total voltage. Therefore, the number of capacitors required in series is:

$n_\text{series} = \frac{1500\,\text{V}}{500\,\text{V}} = 3$

2. Equivalent capacitance of series combination:

The equivalent capacitance of capacitors in series is given by:

$\frac{1}{C_\text{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}$

Since all the capacitors have the same capacitance ($2\,\mu\text{F}$), the equivalent capacitance of the series combination is:

$\frac{1}{C_\text{series}} = \frac{n_\text{series}}{2\,\mu\text{F}} = \frac{3}{2\,\mu\text{F}}$

$C_\text{series} = \frac{2}{3}\,\mu\text{F}$

3. Number of parallel combinations:

To achieve the desired $6\,\mu\text{F}$ capacitance, we need to connect multiple series combinations in parallel. The equivalent capacitance of capacitors in parallel is the sum of their individual capacitances. Let $n_\text{parallel}$ be the number of parallel combinations required. Then:

$C_\text{total} = n_\text{parallel} \times C_\text{series}$

$6\,\mu\text{F} = n_\text{parallel} \times \frac{2}{3}\,\mu\text{F}$

$n_\text{parallel} = \frac{6\,\mu\text{F}}{\frac{2}{3}\,\mu\text{F}} = 9$

4. Total number of capacitors:

The total number of capacitors needed is the product of the number of capacitors in series and the number of parallel combinations:

$n_\text{total} = n_\text{series} \times n_\text{parallel} = 3 \times 9 = 27$

The correct answer is (D) 27.

Answer 2

1. Voltage Requirement:
The circuit requires a potential difference of 1.5 kV = 1500 V. Each capacitor can withstand a potential difference of not more than 500 V. To handle 1500 V, capacitors must be connected in series.
Number of capacitors in series needed in each branch = Required Voltage / Voltage rating of each capacitor = 1500 V / 500 V = 3 capacitors.

2. Capacitance of Series Combination:
When capacitors are connected in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances. For three 2 µF capacitors in series:
$\frac{1}{C_{series}} = \frac{1}{2 \mu F} + \frac{1}{2 \mu F} + \frac{1}{2 \mu F} = \frac{3}{2 \mu F}$
$C_{series} = \frac{2}{3} \mu F$
So, each series branch of 3 capacitors provides a capacitance of $\frac{2}{3} \mu F$ and can withstand 1500 V.

3. Required Capacitance and Parallel Combination:
We need a total capacitance of 6 µF. To increase capacitance, we connect series branches in parallel. In a parallel combination, capacitances add up.
Let 'n' be the number of series branches connected in parallel. The total capacitance $C_{total}$ will be:
$C_{total} = n \times C_{series} = n \times \frac{2}{3} \mu F$
We want $C_{total} = 6 \mu F$.
$n \times \frac{2}{3} \mu F = 6 \mu F$
$n = \frac{6}{\frac{2}{3}} = 6 \times \frac{3}{2} = 9$
So, we need 9 series branches connected in parallel.

4. Total Number of Capacitors:
Number of series branches = 9
Number of capacitors in each series branch = 3
Total number of capacitors = Number of series branches × Number of capacitors in each series branch = 9 × 3 = 27 capacitors.

Therefore, the minimum number of capacitors required is 27.