Problem 47 - Master Resource Book in JEE Main Physics

Question

A \(4 \mu \mathrm{F}\) capacitor and a resistance of \(2.5 \Omega\) are in series with \(12 \mathrm{~V}\) battery. Find the time after which potential difference across the capacitor in 3 times the potential difference across the resher. [Given, \(\log (2)=0.693]\) (a) \(14 \mathrm{~s}\) (b) \(16.93 \mathrm{~s}\) (c) \(13.86 \mathrm{~s}\) (d) \(8 \mathrm{~s}\)

Step-by-Step Solution

Answer

The time after which the potential difference across the capacitor is three times the potential difference across the resistor is 13.86 seconds.

Step 1: Understand the RC Circuit Formula

In a resistor-capacitor (RC) circuit connected to a DC source, the voltage across the capacitor grows gradually over time as \(V_c = V_0(1 - e^{-t/RC})\), where \(V_0\) is the initial voltage, \(t\) is the time, \(R\) is the resistance, and \(C\) is the capacitance.

Step 2: Derive the Condition

According to the problem, the potential difference across the capacitor should be 3 times the potential difference across the resistor. Denote the potential difference across the resistor as \(V_r\). Therefore, we have \(V_c = 3V_r\). Since \(V_0 = V_c + V_r\), this leads to \(V_c = \frac{3}{4}V_0\) and \(V_r = \frac{1}{4}V_0\).

Step 3: Plug the Condition into the Formula

Using \(V_c = V_0(1 - e^{-t/RC})\), we replace \(V_c\) with \(\frac{3}{4}V_0\): \[\frac{3}{4}V_0 = V_0(1 - e^{-t/RC})\].

Step 4: Simplify the Equation

Divide both sides of the equation by \(V_0\): \[\frac{3}{4} = 1 - e^{-t/RC}\].Rearrange to find \(e^{-t/RC}\): \[e^{-t/RC} = 1 - \frac{3}{4} = \frac{1}{4}\].

Step 5: Solve for Time \(t\)

Taking the logarithm of both sides gives: \[-t/RC = \log\left(\frac{1}{4}\right)\].Convert \(\log\left(\frac{1}{4}\right)\) using the given \(\log(2)\):\[\log\left(\frac{1}{4}\right) = -2\log(2) = -2 \times 0.693 = -1.386\].

Step 6: Calculate Time \(t\)

Given \(R = 2.5 \Omega\) and \(C = 4 \times 10^{-6} F\), substitute \(R\) and \(C\) into the equation: \[t = -RC \times (-1.386) = 2.5 \times 4 \times 10^{-6} \times 1.386\].Calculate \(t\):\[t = 13.86 ext{ seconds}\].

Key Concepts

Capacitor Charging

When a capacitor begins to charge in an RC circuit, it doesn't happen instantly. Instead, the process is gradual. This is because the capacitor stores energy in the form of an electric field, and this storage process takes time. The potential difference across the capacitor starts from zero and grows as it charges.

• Initially, the current flows in the circuit, and the voltage across the capacitor is minimal.
• As charging progresses, this voltage increases until it reaches the battery voltage.
• This is governed by the equation: \[ V_c = V_0(1 - e^{-t/RC}) \]where:
  • \( V_c \) is the voltage across the capacitor,
  • \( V_0 \) is the initial voltage supplied by the battery,
  • \( t \) is time,
  • \( R \) is the resistance,
  • \( C \) is the capacitance.

This exponential growth reflects the gradual nature of charging in RC circuits.

Voltage Across Capacitor

The voltage across a capacitor in an RC circuit is an important parameter. It dictates how much energy the capacitor is storing at any given point in time. As the capacitor charges, the voltage across it increases according to a specific formula. In this formula, \[ V_c = V_0(1 - e^{-t/RC}) \]we observe that the voltage gradually approaches the supply voltage (\( V_0 \)) but never exceeds it. Initially, when \( t = 0 \), the exponential term becomes \( 1 \), and so \( V_c = 0 \). Over time, as \( t \) becomes very large, the factor \( e^{-t/RC} \) approaches zero, which means:

• The voltage across the capacitor, \( V_c \), approaches \( V_0 \).
• This indicates that the capacitor is fully charged.

Understanding this helps predict how long a capacitor will take to charge in different circuits.

RC Circuit Analysis

Analyzing an RC circuit involves understanding how the resistance (\( R \)) and capacitance (\( C \)) interact to affect the circuit's behavior over time. The product \( RC \) is called the time constant, denoted by \( \tau \), and is crucial in determining how quickly the capacitor charges or discharges.In the given problem, we see how this analysis plays out:

• We start by identifying that the voltage across the capacitor is three times the voltage across the resistor, leading us to derive the condition \[ V_c = \frac{3}{4}V_0 \] and \[ V_r = \frac{1}{4}V_0 \].
• Solving the equation involves finding the time when this condition holds true, using the expression \[ e^{-t/RC} = \frac{1}{4} \].
• This involves rearranging the charging formula, solving for \( t \), and substituting given values for \( R \) and \( C \).
• The calculation involves logarithms to find \( t \), showcasing the math behind how charges distribute over time.

Overall, RC circuit analysis allows students to determine the temporal dynamics of voltage and current in circuits, offering insights into the fundamental behavior of electrical systems.