Problem 159 - Master Resource Book in JEE Main Physics

Question

To form a composite \(16 \mu \mathrm{F}, 1000 \mathrm{~V}\) capacitor from a supply of identical capacitors marked \(8 \mu \mathrm{F}, 250 \mathrm{~V}\), we required a minimum number of capacitors. [Karnataka CET 2008] (a) 40 (b) 32 (c) 18 (d) 22

Step-by-Step Solution

Answer

32 capacitors are needed to form the composite capacitor.

Step 1: Understanding the Capacitor Values

We need a composite capacitor of \(16 \mu F\) which can withstand \(1000 \mathrm{~V}\). We have a supply of smaller capacitors, each rated at \(8 \mu F\), \(250 \mathrm{~V}\). We need to use these to build the composite capacitor.

Step 2: Voltage Requirements

The composite capacitor must handle up to \(1000 \mathrm{~V}\). Since the smaller capacitors are rated for \(250 \mathrm{~V}\), we need to arrange them in series to multiply the voltage handling capacity. We need \(\frac{1000}{250} = 4\) capacitors in series to reach \(1000 \mathrm{~V}\).

Step 3: Capacitance in Series

When capacitors are connected in series, the total capacitance \(C_s\) is given by the reciprocal formula:\[\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4}\]With each capacitor being \(8 \mu F\), their series equivalent capacitance becomes:\[C_s = \frac{8}{4} = 2 \mu F\]

Step 4: Achieving Desired Capacitance

We need the final capacitance to be \(16 \mu F\). To do this, we must connect multiple series combinations (each \(2 \mu F\)) in parallel. The number of such series combinations needed is \(\frac{16}{2} = 8\).

Step 5: Total Number of Capacitors Needed

Each series group consists of 4 capacitors and we need 8 such groups in total. Therefore, the total number of capacitors required is \(4 \times 8 = 32\).

Key Concepts

Series Capacitors

Capacitors in series are like resistors in parallel. When you connect capacitors in series, you increase the total voltage handling capacity without changing the amount of charge that each can hold. This is particularly useful if you need to handle high voltages, but you only have capacitors with a lower voltage rating.

Here's how it works: when capacitors are connected in series, the reciprocal of the total capacitance is the sum of the reciprocals of each individual capacitance. In mathematical terms, for capacitors C1, C2, ..., and Cn:

• \[\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}\]

This formula means that the total capacitance \(C_s\) is always less than the smallest capacitance in the series. In our example, using four capacitors each with \(8 \mu F\), the total series capacitance becomes \(2 \mu F\).

Remember, to calculate voltage capacity with multiple capacitors in series, you add their voltages. So, if each capacitor handles \(250 \mathrm{~V}\), four series capacitors can handle \(1000 \mathrm{~V}\).

Parallel Capacitors

Now let's talk about parallel capacitors, which is about increasing total capacitance. When capacitors are in parallel, the total capacitance is simply the sum of all individual capacitances. This is very different from what happens in a series set-up. In mathematical terms, for capacitors \( C_1, C_2, \ldots, \) and \( C_n \):

• \[ C_p = C_1 + C_2 + ... + C_n \]

In this way, connecting capacitors in parallel is straightforward—you just add them up! This is ideal for increasing the total amount of stored electrical energy while keeping the voltage rating the same as individual capacitors, which is useful when you don't need to increase voltage capacity.

In our exercise, each series of 4 capacitors (making one series unit \(2 \mu F\)) is repeated 8 times in parallel to achieve the desired \(16 \mu F\). This way, the capacitance increases as needed, while maintaining the required voltage capability.

Composite Capacitance Calculation

Creating a composite capacitor from smaller units requires a strategic combination of both series and parallel connections. The aim is to achieve the desired capacitance and voltage ratings using available resources.

First, ensure each series of capacitors meets the voltage requirement. In our exercise, we need \(1000 \mathrm{~V}\) capability, using individual \(250 \mathrm{~V}\) capacitors. So, four capacitors in series suffice. This arrangement reduces individual \(8 \mu F\) capacitors to a combined \(2 \mu F\).

Next, bring the capacitance up to the required level. To get \(16 \mu F\) total capacitance, you'll place enough of these 2 \(\mu F\) series arrangements in parallel. Multiply the number of required series combinations (\( \frac{16 \mu F}{2 \mu F} = 8\)) by the number of capacitors per series (which is 4), resulting in a total requirement of 32 capacitors.
To sum up,

• Determine the series arrangement for voltage compliance.
• Calculate multiple necessary series in parallel to meet capacitance need.
• Ensure that total capacitors match required specifications.

Understanding this process helps tailor solutions in real-world applications, such as creating electronics components capable of handling specific electrical characteristics efficiently.