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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

Two resistors, \(R_{1}\) and \(R_{2},\) are connected in series across a potential difference, \(\Delta V_{0}\). Express the potential drop across each resistor individually, in terms of these quantities. What is the significance of this arrangement?

You want to make an ohmmeter to measure the resistance of unknown resistors. You have a battery with voltage \(\mathrm{V}_{\mathrm{emf}}=9.00 \mathrm{~V}\), a variable resistor, \(R,\) and an ammeter that measures current on a linear scale from 0 to \(10.0 \mathrm{~mA}\) a) What resistance should the variable resistor have so that the ammeter gives its full-scale (maximum) reading when the ohmmeter is shorted? b) Using the resistance from part (a), what is the unknown resistance if the ammeter reads \(\frac{1}{4}\) of its full scale?

A resistor and a capacitor are connected in series. If a second identical capacitor is connected in series in the same circuit, the time constant for the circuit will a) decrease. b) increase. c) stay the same.

A circuit consists of two \(100 .-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the potential drop across one of the resistors. b) A voltmeter with internal resistance \(10.0 \mathrm{M} \Omega\) is connected in parallel with one of the two resistors in order to measure the potential drop across the resistor. By what percentage will the voltmeter reading deviate from the value you determined in part (a)? (Hint: The difference is rather small so it is helpful to solve algebraically first to avoid a rounding error.)

A capacitor bank is designed to discharge 5.0 J of energy through a \(10.0-\mathrm{k} \Omega\) resistor array in under \(2.0 \mathrm{~ms}\) To what potential difference must the bank be charged, and what must the capacitance of the bank be?
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