A capacitance of 2 uF is required in an electrical circuit across a potential difference of 1 kV. A

Q: Why can't we just use two $1\mu\text{F}$ capacitors in parallel?

While that gives you $2\mu\text{F}$, they would blow up! They can only handle $300\text{ V}$ each, and the circuit is $1000\text{ V}$.

Q: Why do we connect them in series first?

Series connection divides the total voltage among the capacitors, keeping each within its safe $300\text{ V}$ limit.

Q: Does connecting in series reduce the capacitance?

Yes, for $n$ identical capacitors $C$ in series, the equivalent capacitance drops to $C/n$.

Q: Why do we need parallel rows after the series connection?

Because the series connection drops the capacitance too low. Parallel rows add up the capacitance to reach the required $2\mu\text{F}$.

Q: What if we used 5 capacitors in series instead of 4?

It would be safer for voltage ($200\text{ V}$ each), but the equivalent capacitance would be $0.2\mu\text{F}$, requiring 10 rows, totaling 50 capacitors. The question asks for the *minimum* number, so 4 is optimal.

Question

A capacitance of $2\mu	ext{F}$ is required in an electrical circuit across a potential difference of $1	ext{kV}$. A large number of $1\mu	ext{F}$ capacitors are available which can withstand a potential difference of not more than $300	ext{ V}$. The minimum number of capacitors required to achieve this is

Options:

(A) 16

(B) 24

(C) 32

(D) 2

Sigma Percentile · Accepted Answer

Imagine you are an electrical engineer tasked with building a robust circuit. You need a total capacitance of $2\mu	ext{F}$ that can safely operate across a massive $1000	ext{ V}$ potential difference.

However, you look in your inventory and find only $1\mu	ext{F}$ capacitors, and each comes with a strict warning: **Maximum Voltage $300	ext{ V}$**.

If you connect even one of these directly to the $1000	ext{ V}$ source, it will instantly blow up! So, how do we solve this puzzle? We need to strategically combine these smaller, weaker capacitors into a powerful grid.

### Series Connection: Dividing the Voltage

The first problem we must tackle is the voltage constraint. We know from the principles of electrostatics that **when capacitors are connected in series, the total voltage is divided among them**.

To find out how many capacitors we need in a single series row to safely handle $1000	ext{ V}$, we divide the total voltage by the maximum voltage capacity of a single capacitor.

$$n \ge \frac{1000}{300} = 3.33$$

Since we cannot have a fraction of a capacitor, we must round up. Therefore, we need exactly **$4$ capacitors in series**. With 4 capacitors, the $1000	ext{ V}$ is divided equally, meaning each capacitor only experiences $250	ext{ V}$, which is safely below the $300	ext{ V}$ limit.

### Parallel Connection: Boosting the Capacitance

Now that our voltage problem is solved, we face a new issue. **Connecting capacitors in series reduces the overall capacitance**.

For $n$ identical capacitors of capacitance $C$ in series, the equivalent capacitance of that row is given by:

$$C_{row} = \frac{C}{n} = \frac{1\mu	ext{F}}{4} = 0.25\mu	ext{F}$$

Our target is $2\mu	ext{F}$, but our single row only provides $0.25\mu	ext{F}$. To increase the capacitance without affecting the voltage distribution, we must connect multiple such rows in parallel. **In a parallel combination, the equivalent capacitance is simply the sum of the individual capacitances**.

### The Final Grid: Putting it all together

Let's assume we need $m$ such rows connected in parallel. The total capacitance will be $m$ times the capacitance of a single row.

$$m 	imes C_{row} = C_{eq}$$

Substituting our values into the equation:

$$m 	imes 0.25\mu	ext{F} = 2\mu	ext{F}$$

Solving for $m$, we get:

$$m = \frac{2}{0.25} = 8 	ext{ rows}$$

We have successfully designed our grid! We need $8$ rows in parallel, and each row must contain $4$ capacitors in series.

To find the total number of capacitors required, we simply multiply the number of rows by the number of capacitors per row:

$$	ext{Total Capacitors} = m 	imes n = 8 	imes 4 = 32$$

And there you have it! By understanding the fundamental rules of series and parallel combinations, we transformed a pile of weak capacitors into a high-voltage, high-capacity powerhouse.

Select Answer:

Visualized Solution

Target $C_{e q}$ and $V_{t o t a l}$

Voltage Division in Series

Calculating $n$

Number of Capacitors in Series

Equivalent Capacitance of One Row

Capacitance Addition in Parallel

Calculating $m$

Number of Rows in Parallel

Total Number of Capacitors

Conceptual Extension

The Sigma Insight: Combination of Capacitors

Series Connection

Parallel Connection

The Final Grid

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