Question

Many electronics devices can be dangerous even after they are shut off. Consider an \(\mathrm{RC}\) circuit with a \(150 .-\mu \mathrm{F}\) capacitor and a \(1.00-\mathrm{M} \Omega\) resistor connected to a 200.-V power source for a long time and then disconnected and shorted, as shown in the figure. How long will it be until the potential difference across the capacitor drops to below \(50.0 \mathrm{~V} ?\)

Short answer

Answer: It will take approximately \(2.77 \times 10^5\) seconds for the potential difference across the capacitor to drop below 50.0 V.

Step-by-step solution

Write down the given values

We have the following given values: Initial voltage across the capacitor, \(V_0 = 200\ \mathrm{V}\) Voltage across the capacitor at time \(t\), \(V(t) = 50\ \mathrm{V}\) Capacitance, \(C = 150\ \mathrm{\mu F} = 150 \times 10^{-6}\ \mathrm{F}\) Resistance, \(R = 1.00\ \mathrm{M\Omega} = 1.00 \times 10^6\ \Omega\) Our goal is to find the time \(t\) when the voltage across the capacitor drops to 50 V.

Use the voltage formula for a discharging capacitor

The formula for the voltage across a capacitor during the discharge process is given by: \(V(t) = V_0 e^{-\frac{t}{RC}}\) We want to find \(t\) when \(V(t) = 50\ \mathrm{V}\). We can do this by plugging in the given values and solving for \(t\): \(50 = 200 e^{-\frac{t}{(1.00 \times 10^6 \ \Omega)(150 \times 10^{-6}\ \mathrm{F})}}\)

Solve for t

First, we need to isolate \(t\) in the equation. We can do this by dividing both sides by 200 and then taking the natural logarithm of both sides: \(\frac{50}{200} = e^{-\frac{t}{(1.00 \times 10^6 \ \Omega)(150 \times 10^{-6}\ \mathrm{F})}}\) \(\ln{0.25} = -\frac{t}{(1.00 \times 10^6 \ \Omega)(150 \times 10^{-6}\ \mathrm{F})}\) Now we can multiply both sides by the \(RC\) term and then find \(t\): \(t = -\ln{0.25} \times {(1.00 \times 10^6 \ \Omega)(150 \times 10^{-6}\ \mathrm{F})}\) Calculating \(t\): \(t \approx 2.77 \times 10^5\ \mathrm{s}\) Thus, it will take approximately \(2.77 \times 10^5\) seconds for the potential difference across the capacitor to drop below 50.0 V.

Key concepts

Exponential Decay

Understanding the concept of exponential decay is crucial when dealing with circuits involving capacitors. It refers to the process where the quantity of something decreases at a rate proportional to its current value. This can be visualized as a curve that steeply drops off initially and gradually flattens out over time. In the case of an RC circuit, exponential decay describes how the voltage across the capacitor decreases after the power source is removed.

Imagine a ball rolling down a hill; it starts fast but slows as the hill levels out. Similarly, when a capacitor discharges, the voltage drop is swift at first but slows as it approaches zero. It is important to grasp that the 'steepness' of this decay depends on the values of resistance (R) and capacitance (C) of the circuit. A larger resistance or capacitance means the ball 'rolls down' more slowly – the decay of voltage takes longer.

One might find this particularly intuitive by thinking about how certain electronic devices remain powered for a while even after they're turned off. That's the exponential decay of the stored energy in the device's capacitors at work. By understanding exponential decay, students can anticipate the behavior of electrical charges in various circuit components over time.

Capacitor Discharge Equation

The capacitor discharge equation reveals how the voltage across a capacitor changes as it discharges over time. Mathematically, it's expressed as:
\[V(t) = V_0 e^{\frac{-t}{RC}}\]
Where:

• \(V(t)\) is the voltage across the capacitor at time \(t\),
• \(V_0\) is the initial voltage,
• \(e\) is the base of the natural logarithm (approximately equal to 2.71828),
• \(R\) stands for the resistance,
• \(C\) is the capacitance, and
• \(t\) is the time in seconds since the circuit was disconnected from the power source.

It is essential that students can adapt this equation to solve for any of these variables given the others. For example, in the exercise provided, the initial voltage \(V_0\) and the desired voltage \(V(t)\) after some time \(t\) are known, as well as the values of \(R\) and \(C\). Substituting these into the equation allows for the calculation of the time \(t\) it takes for the voltage to fall to a certain level. This equation is a practical tool for predicting the behavior of a capacitor in an electrical circuit, which can be applied in many real-world scenarios like determining the longevity of battery life in electronic devices.

Time Constant

The time constant, symbolized by \( \tau \) (the Greek letter tau), represents a key concept in understanding the discharge of a capacitor in an RC circuit. It is defined as the time it takes for the voltage across the capacitor to drop to about 36.8% of its initial value. The time constant is the product of the resistance and capacitance values within the circuit, given by the formula:
\[\tau = RC\]
In other words, the time constant is the 'speedometer' for the rate of discharge: a larger \(\tau\) value means a slower discharge. It's important for students to recognize that the time constant does not change with the voltage; it is determined solely by the circuit's physical components.

Using the exercise as an example, the time constant of the given RC circuit can be found by multiplying the resistance \(R = 1.00 \times 10^6 \Omega\) by the capacitance \(C = 150 \times 10^{-6} F\), thus:
\[\tau = (1.00 \times 10^6) \times (150 \times 10^{-6}) = 150 \text{s}\]
Understanding and calculating the time constant assists in analyzing how quickly a circuit responds to changes and can be especially relevant in designing electronic timing devices. It helps with grasping the holistic view of the circuit's discharge behavior beyond just the immediate drop in voltage.