Question

A 12.0-mF capacitor is charged to a potential of 50.0 V and then discharged through a 225-\(\Omega\) resistor. How long does it take the capacitor to lose (a) half of its charge and (b) half of its stored energy?

Short answer

(a) 1.88 s for charge, (b) 0.942 s for energy.

Step-by-step solution

Understanding the Problem Statement

We are given a 12.0 mF capacitor charged to 50.0 V, which discharges through a 225 Ω resistor. We need to calculate how long it takes to lose half the charge and half the stored energy.

Identify Key Formulas

For a capacitor-discharge circuit, the charge or voltage decreases exponentially: \[Q(t) = Q_0 e^{-t/RC}\] where \(Q_0\) is the initial charge, \(t\) is the time, \(R\) is the resistance, and \(C\) is the capacitance. Similarly, energy is given by \(E = \frac{1}{2}CV^2\).

Calculate Initial Conditions

First, calculate the initial charge using:\[Q_0 = CV = 12.0 \times 10^{-3} \times 50.0 = 0.6 \text{ C}\].

Determine Time to Lose Half Charge

Use the formula for charge decay, \[\frac{Q_0}{2} = Q_0 e^{-t/RC}\] Solve for \(t\):\[\frac{1}{2} = e^{-t/RC}\]Simplifying, we have:\[t = RC \ln(2)\]Calculate:\[t = 225 \times 12.0 \times 10^{-3} \times \ln(2) \approx 1.88 \text{ s}\].

Calculate Initial Energy

Use the energy formula: \[E_0 = \frac{1}{2}CV^2 = \frac{1}{2} \times 12.0 \times 10^{-3} \times (50.0)^2 = 15.0 \text{ J}\].

Determine Time to Lose Half Energy

Since energy relates to voltage as \[E = \frac{1}{2}CV^2\], replace voltage in the equation. Half of the energy means \[E = \frac{E_0}{2} = \frac{1}{4}CV_0^2\] (since energy depends on \(V^2\)).This corresponds to voltage \(V = V_0/\sqrt{2}\) using \[V(t) = V_0 e^{-t/RC}\]. Now solve:\[\frac{1}{\sqrt{2}} = e^{-t/RC}\]and \[t = RC \ln(\sqrt{2}) = \frac{RC}{2} \ln(2)\]Calculate:\[t \approx 0.942 \text{ s}\].

Key concepts

Exponential Decay

Consider how a capacitor discharges in a circuit, specifically focusing on the concept of exponential decay. In such a scenario, the charge on a capacitor decreases exponentially over time. This means that the rate at which the charge decreases is proportional to the amount of charge currently remaining.

Exponential decay can be described mathematically by the formula:

• \( Q(t) = Q_0 e^{-t/RC} \)

Here, \(Q_0\) is the initial charge, \(t\) is the time that has elapsed, \(R\) is the resistance, and \(C\) is the capacitance. The main takeaway here is that the charge reduces by a fixed percentage over each time interval.

This characteristic behavior allows us to predict precisely how long it takes for the charge to reach a certain percentage of its original value, such as when it reduces to half.

RC Circuit

An RC circuit is a simple and common electrical circuit involving a resistor (R) and a capacitor (C). The resistor controls the flow of charge, while the capacitor stores electrical energy. Together, they create a dynamic circuit where the charge flows from the capacitor through the resistor.

When a charged capacitor is connected to the resistor, it begins to discharge its stored energy. The resistor slows down this process, creating an exponential decrease in voltage and charge over time.

The time constant \( \tau \) of an RC circuit is a key parameter, calculated as the product of resistance and capacitance:

• \( \tau = RC \)

This constant represents the time it takes for the charge to reduce to approximately 37% of its initial value. Understanding this interplay is crucial to solving problems involving the discharge of capacitors in RC circuits.

Stored Energy in Capacitors

Capacitors are devices that store energy in an electric field between their plates. The amount of energy stored can be determined using the formula:

• \( E = \frac{1}{2}CV^2 \)

where \(C\) is capacitance and \(V\) is voltage.

Initially, the energy stored is at its maximum when the capacitor is fully charged. As the capacitor discharges, the voltage across its plates decreases and so does the stored energy.

• Because the energy stored is proportional to the square of the voltage, a small decrease in voltage results in a larger percentage decrease in energy.
• This exponential decrease in energy parallels the exponential decay of charge in the RC circuit, and it's crucial for understanding how quickly a capacitor loses energy.

Being able to calculate the time it takes for a capacitor to lose half of its stored energy involves understanding these fundamental relationships.

Electrical Resistance

Electrical resistance is a measure of how much an object resists the flow of electric current. It is measured in ohms (\(\Omega\)), and it's crucial in determining how fast a capacitor discharges in an RC circuit.

The resistor provides opposition to the flow of electrons, thereby controlling the discharge time of the capacitor. A higher resistance means slower discharge because it provides a greater opposition to current flow.

This opposition is beautifully illustrated in our key equations for an RC circuit, where resistance directly affects both the time constant \( \tau \) and the rate of charge decay. Key insights include:

• The larger the resistance, the longer it takes for the capacitor to discharge.
• Understanding resistance helps in evaluating the discharge time formulas, such as \( t = RC \ln(2) \) for determining how long it takes to lose half of the charge.

By grasping how resistance interacts with capacitance, one can predict how quickly or slowly a capacitor will discharge in an electrical circuit.