Problem 136 - Master Resource Book in JEE Main Physics

Question

Let \(C\) be the capacitance of a capacitor discharging through a resistor \(R\). Suppose \(t_{1}\), is the time taken for the energy stored in the capacitor to reduce to half its initial value and \(t_{2}\) is the time taken for the charge to reduce to one-fourth its initial value. Then, the ratio \(\underline{t_{1}}\) will be [AIEEE 2010] \(t_{2}\) (a) 1 (b) \(\frac{1}{2}\) (c) \(\frac{1}{4}\) (d) 2

Step-by-Step Solution

Answer

The ratio \( \frac{t_{1}}{t_{2}} \) is \( \frac{1}{4} \), which corresponds to option (c).

Step 1: Determine the expressions for energy and charge

The energy stored in a capacitor at any time is given by \( U = \frac{1}{2}CV^2 \). The charge on the capacitor is \( Q = CV \), leading to an expression for voltage \( V = \frac{Q}{C} \). Energy can be rewritten as \( U = \frac{1}{2}C\left(\frac{Q}{C}\right)^2 = \frac{Q^2}{2C} \).

Step 2: Determine time for energy to halve

To find the time for energy to halve, \( U_{t_1} = \frac{U_0}{2} \), where \( U_0 \) is the initial energy. Substituting the energy equation: \[ \frac{Q^2}{2C} = \frac{Q_0^2}{4C} \Rightarrow Q^2 = \frac{Q_0^2}{2} \Rightarrow Q = Q_0 \frac{1}{\sqrt{2}}. \] Using the relation for charge decay in an RC circuit, \( Q = Q_0 e^{-\frac{t_1}{RC}} \), set \( Q = Q_0 \frac{1}{\sqrt{2}} \) and solve for \( t_1 \).

Step 3: Calculate expression for t1

From the charge expression, \( e^{-\frac{t_1}{RC}} = \frac{1}{\sqrt{2}} \), take the natural log of both sides to get \( -\frac{t_1}{RC} = \ln\left(\frac{1}{\sqrt{2}}\right) = -\frac{1}{2}\ln(2) \). Solving gives \( t_1 = \frac{RC}{2}\ln(2) \).

Step 4: Determine time for charge to quarter

For charge to reduce to a quarter, \( Q_{t_2} = \frac{Q_0}{4} \). Set up \( Q = Q_0 e^{-\frac{t_2}{RC}} \), so \( e^{-\frac{t_2}{RC}} = \frac{1}{4} \). Take the natural log to solve for \( t_2 \).

Step 5: Calculate expression for t2

Taking the natural log, \( -\frac{t_2}{RC} = \ln\left(\frac{1}{4}\right) = -2\ln(2) \). This gives \( t_2 = 2RC\ln(2) \).

Step 6: Find the ratio \( \frac{t_1}{t_2} \)

Divide the expression for \( t_1 \) by \( t_2 \): \[ \frac{t_1}{t_2} = \frac{\frac{RC}{2}\ln(2)}{2RC\ln(2)} = \frac{1}{4}. \]

Key Concepts

Capacitance

In electronics, capacitance is a key property of a capacitor, which allows it to store and release electrical energy. It is defined as the amount of electrical charge a capacitor can store per unit voltage. The unit of capacitance is the farad (F).

Capacitors come in various types, and each has a specific capacitance value given by the formula:

• \( C = \frac{Q}{V} \)

where \( Q \) is the charge stored on the capacitor and \( V \) is the voltage across its plates. As the voltage increases, more charge gets stored in the capacitor.

Understanding capacitance is crucial for analyzing RC circuits. In these circuits, resistors (R) and capacitors (C) work together to manage the flow of current, control the timing, and smooth out the fluctuations in electronic signals.

Charge Decay

Charge decay relates to how the charge stored in a capacitor diminishes over time when connected to a resistor. In an RC circuit, the charge on a capacitor decreases exponentially from its initial value.

As the circuit discharges, the charge \( Q \) at any time \( t \) is given by:

• \( Q = Q_0 e^{-\frac{t}{RC}} \)

where \( Q_0 \) is the initial charge, \( R \) is the resistance, and \( C \) is the capacitance.

In this context, understanding the rate at which charge decays helps in predicting how quickly the capacitor releases its stored energy. Charge decay is a critical factor in designing circuits that require precise timing elements, such as in oscillators and timers.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. This characteristic is common in RC circuits, particularly in the context of how voltage, charge, or current in a capacitor diminishes over time.

For example, if you have a quantity \( Q \) at the initial time, it decreases exponentially as given by:

• \( Q = Q_0 e^{-kt} \)

where \( k \) is the decay constant. In an RC circuit, \( k \) corresponds to \( \frac{1}{RC} \).

This exponential behavior is crucial to understand how systems react to changes. For example, exponential decay ensures that the capacitor does not discharge instantly, allowing for gradual changes that are useful in filtering and timing applications.

Logarithmic Functions

Logarithmic functions are vital in the analysis of RC circuits, especially when examining time constants and decay processes. In the context of this problem, natural logarithms (\( \ln \)) are used to solve equations involving exponential decay.

Consider the time it takes for a certain process, like charge reduction, to occur:

• Given the equation \( Q = Q_0 e^{-\frac{t}{RC}} \), solving for time involves logarithms.

For example, if you need to find the time \( t \) at which the charge reduces by a certain fraction, you take the natural log of both sides:

• \( -\frac{t}{RC} = \ln\left(\frac{Q}{Q_0}\right) \)

This relationship is especially useful in understanding how different components in a circuit interact over time, helping design efficient electronic systems.