Problem 129 - Master Resource Book in JEE Main Physics

Question

Combination of two identical capacitors, a resistor \(R\) and a DC voltage source of voltage \(6 \mathrm{~V}\) is used in an experiment on \(C-R\) circuit. It is found that for a parallel combination of the capacitor the time in which the voltage of the fully charged combination reduces to half its original voltage is \(10 \mathrm{~s}\). For series combination, the time needed for reducing the voltage of the fully charged series combination by half is (a) \(20 \mathrm{~s}\) (b) \(10 \mathrm{~s}\) (c) \(5 \mathrm{~s}\) (d) \(2.5 \mathrm{~s}\)

Step-by-Step Solution

Answer

The correct answer is (d) 2.5 s.

Step 1: Understanding the problem

We are given a circuit with capacitors that can be combined in either series or parallel, and a resistor and DC voltage source. We need to determine the time it takes for the voltage in a series capacitor configuration to halve, given that it takes 10 seconds for a parallel configuration.

Step 2: Identifying the formula for voltage decay

In a RC circuit, the voltage decay over time is described by the equation \( V(t) = V_0 e^{-\frac{t}{RC}} \), where \( V_0 \) is the initial voltage, \( R \) the resistance, and \( C \) the equivalent capacitance of the configuration.

Step 3: Calculating time for parallel configuration

For capacitors in parallel, the total capacitance \( C_p = 2C \). We are given that the time \( t_p = 10 \) seconds for the voltage to reach half, which means \( \frac{1}{2} = e^{-\frac{10}{R \cdot 2C}} \).

Step 4: Simplifying using common exponential form

Using the property \( e^{-x} = \frac{1}{2} \) implies \( x = \ln(2) \). Hence, from \( t_p \), we get \( \frac{10}{R \cdot 2C} = \ln(2) \).

Step 5: Determining time for series configuration

For capacitors in series, the total capacitance \( C_s = \frac{C}{2} \). Thus, the time \( t_s \) for voltage to halve is given by \( \frac{t_s}{R \cdot \frac{C}{2}} = \ln(2) \).

Step 6: Relating series time to parallel time

From the equation \( \frac{10}{R \cdot 2C} = \ln(2) \), we know that \( 10 = R \cdot 2C \cdot \ln(2) \). For the series case, \( t_s = R \cdot \frac{C}{2} \cdot \ln(2) \cdot 2 \), which simplifies to \( t_s = 2.5 \).

Step 7: Conclusion

The time needed for the voltage of the fully charged series combination to reduce to half its original voltage is 2.5 seconds.

Key Concepts

Capacitors in Series and Parallel

In electrical circuits, capacitors can be combined in two main ways: series or parallel. These configurations affect how capacitors store and release energy, which is crucial for various applications.
When capacitors are connected in parallel, each capacitor experiences the same voltage. The total or equivalent capacitance in a parallel configuration is the sum of the individual capacitances.

• Formula: For two capacitors in parallel, the total capacitance is: \( C_p = C_1 + C_2 \).

This means the overall capacitance increases with each additional capacitor, allowing more charge storage.In a series configuration, capacitors are connected end to end, and the same charge flows through each. The challenge here is that the total capacitance is less than any individual capacitor’s capacitance.

• Formula: For two capacitors in series, the equivalent capacitance is: \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \).

This configuration effectively increases the circuit's ability to handle higher voltages but with a reduced total capacitance.

Exponential Voltage Decay

In RC (resistor-capacitor) circuits, the voltage across a capacitor decreases exponentially over time during a discharge process. This behavior is governed by a simple yet crucial formula:

• \( V(t) = V_0 e^{-\frac{t}{RC}} \), where:
• \( V(t) \) is the voltage at time \( t \),
• \( V_0 \) is the initial voltage,
• \( R \) is the resistance,
• \( C \) is the equivalent capacitance,
• \( e \) is the base of the natural logarithm.

This formula illustrates that the voltage doesn't drop linearly, but rather decreases at a rate that initially is fast and then slows down as it approaches zero.
The time it takes for the voltage to reduce to half of its initial value is known as the "time constant." The significance of the time constant \( \tau \) is often highlighted, as it determines the speed of the discharge:

• For a given RC circuit, the time constant is \( \tau = RC \).
• This time constant is fundamental when analyzing transient responses of circuits.

Understanding exponential decay is essential in predicting how quickly a capacitor can release its stored energy.

Equivalent Capacitance Calculation

Calculating equivalent capacitance is a foundational skill in understanding how capacitors impact circuit behavior. Whether in series or parallel, this calculation helps determine how capacitors will store and manage charge.For capacitors in parallel, the calculation is straightforward as the individual capacitances simply add up, reflecting a direct increase in the system's charge storage capacity:

• Total capacitance \( C_p = C_1 + C_2 \).

In contrast, capacitors in series require a different approach, where the inverse of the total is the sum of the inverses of the individual capacitances. This results in a total capacitance lesser than any of the components:

• The series capacitance formula: \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \).

For example, if you have two identical capacitors, the equivalent capacitance is halved in series and doubled in parallel. Knowing how to calculate equivalent capacitance allows you to design more efficient RC circuits.