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Chapter 26: Problem 56

Many electronics devices can be dangerous even after they are shut off. Consider an RC circuit with a \(150 .-\mu \mathrm{F}\) capacitor and a \(1.00-\mathrm{M} \Omega\) resistor connected to a \(200 .-\mathrm{V}\) 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

Expert verified
Answer: Approximately 207.9 seconds.

Step by step solution

01

Data given

We have the following data given for the RC circuit: 1. Capacitance of the capacitor, \(C = 150 \mu \text{F} = 150 \times 10^{-6} \text{F}\) 2. Resistance of the resistor, \(R = 1.00 \text{M}\Omega = 1.00 \times 10^6 \Omega\) 3. Initial voltage across the capacitor, \(V_0 = 200 \text{V}\) 4. Final voltage across the capacitor, \(V = 50 \text{V}\)
02

Calculating the time constant

We need to calculate the time constant (\(\tau\)) of the RC circuit, given by the formula: $$\tau = RC$$ Substitute the given values of \(R\) and \(C\) in the formula: $$\tau = (1.00 \times 10^6 \Omega)(150 \times 10^{-6} \text{F}) = 150 \text{s}$$
03

Using the voltage formula

To find the time (\(t\)) it takes for the voltage across the capacitor to drop below 50V, we will use the formula for the voltage across a discharging capacitor in an RC circuit: $$V = V_0 e^{-t/\tau}$$ Here, \(V_0\) is the initial voltage, \(V\) is the final voltage and \(\tau\) is the time constant.
04

Solving for time

Now we need to solve for \(t\): $$V = V_0 e^{-t/\tau}$$ $$50 = 200 e^{-t/150}$$ Divide by 200: $$\frac{1}{4} = e^{-t/150}$$ Take the natural logarithm of both sides: $$-\ln(4) = -\frac{t}{150}$$ Multiply by (-1) and then by 150 to solve for \(t\): $$t = 150\ln(4)$$ Approximately, \(t \approx 207.9 \text{s}\)
05

Conclusion

It will take approximately 207.9 seconds for the potential difference across the capacitor to drop below 50V.

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Most popular questions from this chapter

An ammeter with an internal resistance of \(53 \Omega\) measures a current of \(5.25 \mathrm{~mA}\) in a circuit containing a battery and a total resistance of \(1130 \Omega\). The insertion of the ammeter alters the resistance of the circuit, and thus the measurement does not give the actual value of the current in the circuit without the ammeter. Determine the actual value of the current.

A circuit consists of two \(1.00-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the current flowing through each resistor. b) A student trying to measure the current flowing through one of the resistors inadvertently connects an ammeter in parallel with that resistor rather than in series with it. How much current will flow through the ammeter, assuming that it has an internal resistance of \(1.0 \Omega ?\)

How long will it take for the current in a circuit to drop from its initial value to \(1.50 \mathrm{~mA}\) if the circuit contains two \(3.8-\mu \mathrm{F}\) capacitors that are initially uncharged, two \(2.2-\mathrm{k} \Omega\) resistors, and a \(12.0-\mathrm{V}\) battery all connected in series?

A battery has \(V_{\text {emf }}=12.0 \mathrm{~V}\) and internal resistance \(r=1.00 \Omega\). What resistance, \(R,\) can be put across the battery to extract \(10.0 \mathrm{~W}\) of power from it?

A parallel plate capacitor with \(C=0.050 \mu \mathrm{F}\) has a separation between its plates of \(d=50.0 \mu \mathrm{m} .\) The dielectric that fills the space between the plates has dielectric constant \(\kappa=2.5\) and resistivity \(\rho=4.0 \cdot 10^{12} \Omega \mathrm{m} .\) What is the time constant for this capacitor? (Hint: First calculate the area of the plates for the given \(C\) and \(\kappa\), and then determine the resistance of the dielectric between the plates.)
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