Question

A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40 M\(\Omega\). After a time of 4.00 s the voltmeter reads 3.0 V. What are (a) the capacitance and (b) the time constant of the circuit?

Short answer

Capacitance is approximately \( 8.47 \times 10^{-7} \text{ F} \) and the time constant is 2.88 seconds.

Step-by-step solution

Understand the problem

We are given a circuit with a capacitor initially charged to 12.0 V, which discharges through a voltmeter with an internal resistance of 3.40 MΩ. The voltmeter reads 3.0 V after 4.00 s. We need to find the capacitance of the capacitor and the time constant of the circuit.

Formula for Voltage in RC Circuit

The voltage across a capacitor in a discharging RC circuit is given by the formula:\[ V(t) = V_0 \times e^{-t/RC} \]where \( V_0 \) is the initial voltage, \( V(t) \) is the voltage at time \( t \), \( R \) is the resistance, and \( C \) is the capacitance.

Rearrange the formula to solve for RC

We can rearrange the formula to solve for \( RC \):\[ RC = -\frac{t}{\ln(V(t)/V_0)} \]Substitute \( V(t) = 3.0 \) V, \( V_0 = 12.0 \) V, and \( t = 4.00 \) s.

Calculate the Time Constant (RC)

Substitute the values into the formula:\[ RC = -\frac{4.00}{\ln(3.0/12.0)} \]Calculate the natural logarithm and solve:\[ \ln(3/12) = \ln(0.25) \approx -1.386 \]\[ RC \approx -\frac{4.00}{-1.386} \approx 2.88 \text{ seconds} \]

Use Resistance to Find Capacitance

We know \( RC = 2.88 \) seconds, and the resistance \( R = 3.40 \times 10^6 \) Ω. Solve for \( C \):\[ C = \frac{2.88}{3.40 \times 10^6} \approx 8.47 \times 10^{-7} \text{ F} \]

Summary of Results

The capacitance of the capacitor is approximately \( 8.47 \times 10^{-7} \text{ F} \), and the time constant of the circuit is approximately 2.88 seconds.

Key concepts

Capacitance

Capacitance is a fundamental property of capacitors that defines their ability to store electrical charge. In simpler terms, capacitance represents how much charge a capacitor can hold for a given voltage across its terminals. In the context of our exercise, the capacitor in question is initially charged to 12.0 V. After being connected to the voltmeter with an internal resistance, the challenge is to determine how much electrical energy it can store given this new configuration.

The relationship between charge (\( Q \)) and voltage (\( V \)) in a capacitor is expressed as \( Q = C \times V \), where \( C \) is the capacitance. Therefore, when we connect this capacitor in an RC circuit, the time it takes to discharge significantly depends on this capacity. By analyzing how the voltage changes over time, we can calculate the capacitance using the changes in voltage and known resistance. This exercise calculated \( C \approx 8.47 \times 10^{-7} \) F, indicating the capacitor's ability to store charge under given conditions.

Time Constant

The time constant (\( \tau \)) in an RC circuit is a key measure of how quickly the circuit responds to changes in voltage. It is essentially the time required for a capacitor to discharge to about 37% of its initial value or to charge up to about 63% of its maximum capacity. This time constant plays a crucial role in determining how long a circuit will take to either charge or discharge.

In mathematical terms, the time constant is the product of resistance (\( R \)) and capacitance (\( C \)) of the circuit: \( \tau = R \cdot C \). In the exercise, this is computed as approximately 2.88 seconds. It tells us that in 2.88 seconds, the voltage across the capacitor in our circuit will have dropped significantly from its initial value. Understanding this concept helps students predict how circuits behave over time and adjust their designs or expectations accordingly.

Voltmeter Resistance

In the context of our circuit, voltmeter resistance refers to the internal resistance present within the voltmeter that is part of the measurement process. When connected across a capacitor, the voltmeter acts as an additional resistor in the circuit—here, with a resistance of 3.40 MΩ.

This internal resistance is crucial because it directly impacts the rate at which the capacitor discharges. A higher resistance will slow down the discharge, leading the voltmeter to show a slower decrease in voltage readings over time. Conversely, a lower resistance would result in a quicker discharge and a more rapid drop in voltage readings. Hence, the voltmeter not only measures the voltage but also influences the circuit's behavior by being part of it. The correct consideration of this resistance is essential for accurate calculations and understanding of the RC circuit's dynamics.