Question

During a physics demonstration, a fully charged \(90.0-\mu \mathrm{F}\) capacitor is discharged through a \(60.0-\Omega\) resistor. How long will it take for the capacitor to lose \(80.0 \%\) of its initial energy?

Short answer

Answer: It takes approximately 52.6 seconds for the capacitor to lose 80% of its initial energy.

Step-by-step solution

Find the initial energy stored in the capacitor

First, let's find the initial energy stored in the capacitor using the formula: \(E = \dfrac{1}{2}CV^2\) where \(E\) is the energy, \(C\) is the capacitance, and \(V\) is the voltage. Given that the capacitor is fully charged, we can assume that its initial voltage is equal to its maximum voltage \(V_0\). In the problem, no initial voltage is given, but we can rewrite this equation in terms of the final remaining energy of 20%. Since we don't have the initial voltage value, we will denote the energy as a percentage of the initial energy. Let's denote the initial energy as \(E_0\), and the remaining energy as \(E_f\). We can write the equation for \(E_f\) in terms of \(E_0\): \(E_f = 0.2 E_0\) \(E_0 = \dfrac{1}{2} C V_0^2\)

Find the voltage across the capacitor during the discharge process

The voltage across a capacitor during discharge through a resistor is given by: \(V(t) = V_0 e^{-\frac{t}{RC}}\) where \(V(t)\) is the voltage at time \(t\), \(R\) is the resistance, \(C\) the capacitance, and \(V_0\) the initial voltage.

Find the energy stored in the capacitor at any time

At any time \(t\), the energy stored in the capacitor is given by: \(E(t) = \dfrac{1}{2} C V(t)^2\) To find the energy at the time when the capacitor has lost 80% of its initial energy, we can substitute the discharge voltage equation from Step 2 into this equation: \(E(t) = \dfrac{1}{2} C (V_0 e^{-\frac{t}{RC}})^2\)

Set the final energy equal to 20% of the initial energy

We know that the final energy must be 20% of the initial energy, so we can set the equation from Step 3 equal to \(0.2 E_0\): \(0.2 E_0 = \dfrac{1}{2} C (V_0 e^{-\frac{t}{RC}})^2\) Now substitute the initial energy expression from Step 1 into the above equation: \(0.2 (\dfrac{1}{2} C V_0^2) = \dfrac{1}{2} C (V_0 e^{-\frac{t}{RC}})^2\)

Solve for time \(t\)

First, note that the terms \(\dfrac{1}{2}C\) cancel each other out. So, we are now left with the equation: \(0.2 V_0^2 = V_0^2 e^{-\frac{2t}{RC}}\) Divide both sides by \(V_0^2\): \(0.2 = e^{-\frac{2t}{RC}}\) Now, take the natural logarithm of both sides: \(\ln(0.2) = -\frac{2t}{RC}\) Finally, solve for \(t\): \(t = -\dfrac{RC}{2} \ln(0.2)\)

Plug in the values of \(R\) and \(C\) to find the time \(t\)

Using the given values for the resistance and capacitance, we can find the time it takes for the capacitor to lose 80% of its initial energy: \(t = -\dfrac{(60.0\,\Omega)(90.0 \times 10^{-6}\,\mathrm{F})}{2} \ln(0.2)\) Calculate the value of \(t\): \(t \approx 52.6\) seconds Therefore, it will take approximately \(52.6\) seconds for the capacitor to lose \(80.0 \%\) of its initial energy.

Key concepts

Capacitor Energy Loss

In the realm of electrical circuits, a capacitor's ability to store energy is pivotal for various applications. Capacitors can discharge, slowly releasing the stored energy through other components.

When a capacitor discharges through a resistor, the energy loss is not a linear decline but decreases according to a curve known as an exponential decay. Principally, the energy within the capacitor is given by the formula: \(E = \dfrac{1}{2}CV^2\), where \(E\) is the energy in joules, \(C\) is the capacitance in farads, and \(V\) is the voltage across the capacitor in volts.

The initial energy is at its peak when the capacitor is fully charged, and as it discharges, this energy diminishes. To find out when a capacitor has lost a certain percentage of its energy, we must take into account both the capacitor's capacitance and the resistance through which it is discharging. The rate at which the energy decreases depends on the time constant of the circuit, defined as \(RC\), with \(R\) being the resistance in ohms.

Knowing how to calculate the duration for a capacitor to lose a specific portion of its energy can be invaluable for timing applications and understanding the dynamics of electrical circuits.

Resistor-Capacitor (RC) Circuit

A resistor-capacitor (RC) circuit is one of the simplest and most instructive types of electric circuits used to examine the behavior of capacitors and resistors in tandem. This circuit can filter signals, delay currents, and even serve as a simple memory element in electronics.

The time constant, denoted \(\tau\), is a crucial concept in RC circuits, defined mathematically as \(\tau = RC\). It represents the time it takes for the voltage across the capacitor to either charge up to approximately 63.2% of the difference between the initial and final value, or to discharge down to about 36.8% of its initial value.

As components, the resistor controls the rate at which the capacitor charges or discharges, while the capacitor stores and releases energy. Understanding RC circuits is essential as they mimic many biological and physical systems' responses and are a cornerstone in electronics and signal processing.

Exponential Decay in Circuits

Exponential decay is a fundamental concept when observing capacitors discharging over time in an RC circuit. Rather than releasing energy at a steady rate, the capacitor's voltage and stored energy decrease exponentially.

The voltage across a discharging capacitor at any time \(t\) is described by the equation \(V(t) = V_0 e^{-\frac{t}{RC}}\), where \(V_0\) is the initial voltage, and \(e\) is Euler's number, approximately equal to 2.71828. This tells us that the voltage isn't halved every constant interval; rather, it takes successively longer intervals to halve each time.

In educational settings, applying the concept of exponential decay to circuits aids in comprehending how different systems return to equilibrium. Whether in oscillation dampening or radioactive decay, the exponential nature of these processes is a key aspect of physical phenomena. By grasping exponential decay, students can predict and calculate how long it would take for a system to reduce to a certain energy level, an invaluable skill in circuit design and analysis.