Problem 82 - Master Resource Book in JEE Main Physics

Question

A \(10 \mu \mathrm{F}\) capacitor and a \(20 \mu \mathrm{F}\) capacitor are connected in series across \(200 \mathrm{~V}\) supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor? (a) \(\frac{800}{9} \mathrm{~V}\) (b) \(\frac{800}{3} \mathrm{~V}\) (c) \(400 \mathrm{~V}\) (d) \(200 \mathrm{~V}\)

Step-by-Step Solution

Answer

(a) \( \frac{800}{9} \mathrm{~V} \)

Step 1: Calculate Equivalent Capacitance in Series

When capacitors are connected in series, the total capacitance \(C_t\) can be calculated using the formula: \[ \frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} \] where \(C_1 = 10 \mu \mathrm{F}\) and \(C_2 = 20 \mu \mathrm{F}\). Thus, \[ \frac{1}{C_t} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20} \] Therefore, the total capacitance \(C_t\) is: \[ C_t = \frac{20}{3} \mu \mathrm{F} \]

Step 2: Find Total Charge on Capacitors

The total charge \(Q\) on the capacitors is given by the equation \(Q = C_t \cdot V\), where \(V = 200 \mathrm{~V}\). Substitution gives: \[ Q = \frac{20}{3} \times 200 = \frac{4000}{3} \mu \mathrm{C} \] This is the charge on each capacitor when connected in series.

Step 3: Calculate Initial Potential Difference Across Each Capacitor

For capacitors in series, the voltage across each can be found using \(V = \frac{Q}{C}\). For the 10 \(\mu\mathrm{F}\) capacitor: \[ V_1 = \frac{4000/3}{10} = \frac{400}{3} \mathrm{~V} \] and for the 20 \(\mu\mathrm{F}\) capacitor: \[ V_2 = \frac{4000/3}{20} = \frac{200}{3} \mathrm{~V} \]

Step 4: Analyze Configuration After Reconnection

When the capacitors are reconnected with matched plates, i.e., positive to positive and negative to negative, they essentially form a parallel connection with combined charge and the new potential distribution needs to be calculated.

Step 5: Formulate Equation for Charge Conservation

The total charge remains the same as when connected in series. If \(V_f\) is the final potential difference across each, \[ Q_1 + Q_2 = (C_1 + C_2)V_f \] where \(Q_1 = 10V_f\) and \(Q_2 = 20V_f\). Since total \(Q = \frac{4000}{3} \), \[ 10V_f + 20V_f = \frac{4000}{3} \]

Step 6: Solve for Potential Difference After Reconnection

Solving for \(V_f\) from the above equation, \[ 30V_f = \frac{4000}{3} \] \[ V_f = \frac{4000}{90} = \frac{800}{9} \mathrm{~V} \] Thus, the potential difference across each capacitor is \( \frac{800}{9} \mathrm{~V} \).

Key Concepts

Equivalent Capacitance

In a series circuit of capacitors, the overall effect is different than when they are connected in parallel. When capacitors are connected in series, the total or equivalent capacitance is less than any single capacitance in the circuit. This is calculated using the formula:

• \( \frac{1}{C_t} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots \)

This equation expresses that the reciprocal of the total capacitance equals the sum of the reciprocals of the individual capacitances. For example, a 10 \( \mu \mathrm{F} \) capacitor and a 20 \( \mu \mathrm{F} \) capacitor in series have a combined effect lower than 10 \( \mu \mathrm{F} \).
By substituting the values into the formula, we find the equivalent capacitance \( C_t \) to be \( \frac{20}{3} \mu \mathrm{F} \). This smaller capacitance shows that series capacitors store less charge for the same voltage as compared to if they were alone or in parallel.

Charge Conservation

Charge conservation is a crucial principle in understanding capacitor systems. This principle states that the total charge in an isolated system remains constant regardless of the processes within the system.
For the series connection of our capacitors, the total charge \( Q \) is calculated using the formula:

• \( Q = C_t \cdot V \)

Given the equivalent capacitance \( C_t = \frac{20}{3} \mu \mathrm{F} \) and a potential difference of 200 V, the total charge stored is \( \frac{4000}{3} \mu \mathrm{C} \).
Even when these capacitors are disconnected from the source and reconnected with their positive plates together and their negative plates together, charge conservation ensures that this total charge remains \( \frac{4000}{3} \mu \mathrm{C} \). The capacitors will redistribute this charge among themselves.

Potential Difference Calculation

Calculating the potential difference across each capacitor helps us understand voltage distribution in the system. Initially, when capacitors are connected in series, each capacitor bears a portion of the total voltage based on its capacitance.

• For a 10 \( \mu \mathrm{F} \) capacitor, the initial potential difference is \( \frac{400}{3} \mathrm{~V} \).
• For a 20 \( \mu \mathrm{F} \) capacitor, it is \( \frac{200}{3} \mathrm{~V} \).

This is calculated using the formula:

• \( V = \frac{Q}{C} \)

Once the capacitors are reconnected, the voltage needs recalculation. Despite this reconnection, the magnitude of charge and thus the overall potential difference is adjusted according to the values of the capacitances involved. Hence, both capacitors now have a common potential \( V_f \) across them which satisfies the charge conservation equation.

Capacitor Reconnection Analysis

Reconnecting capacitors after an initial charge distribution can change their charge-sharing arrangement. In this exercise, when we reconnect the charged capacitors with positive to positive and negative to negative, it is equivalent to connecting them in parallel.This new configuration requires us to calculate the potential across each capacitor, given the static charge \( Q \).
By employing the equation for charge conservation:

• \( 10V_f + 20V_f = \frac{4000}{3} \)

we solve for the common potential \( V_f \).
The total charge distribution remains constant, but through recalculating, we find that each capacitor ends up having a potential difference of \( \frac{800}{9} \mathrm{~V} \). This demonstrates how the reconnection affects the internal dynamics of the system without any external influence.