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Chapter 18: Problem 87

Capacitors are used in many applications where one needs to supply a short burst of relatively large current. A 100.0 - \(\mu\) F capacitor in an electronic flash lamp supplies a burst of current that dissipates \(20.0 \mathrm{J}\) of energy (as light and heat) in the lamp. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge? (c) Approximately what is the resistance of the lamp if the current reaches \(5.0 \%\) of its original value in $2.0 \mathrm{ms} ?$

Short Answer

Expert verified
Question: A capacitor of 100.0 - μμF is used to supply energy to turn on a small lamp. When the lamp turns on, the capacitor dissipates 20.0J of energy across the lamp. The current in the lamp reaches 5% of its original value after 2 ms. Calculate: a) The initial potential difference across the capacitor. b) The initial charge on the capacitor. c) The approximate resistance of the lamp. Answer: a) The initial potential difference across the capacitor is 1000V. b) The initial charge on the capacitor is 100.0 x 10^-3 C. c) The approximate resistance of the lamp is 190.5Ω.

Step by step solution

01

Calculate the initial voltage

To find the initial potential difference (voltage) across the capacitor, use the formula for the energy stored in a capacitor: \(E = \frac{1}{2}CV^2\) where \(E\) is the energy, \(C\) is the capacitance, and \(V\) is the voltage. We have \(E = 20.0J\) and \(C = 100.0 \times 10^{-6}F\). Solving for \(V\), we get: \(V = \sqrt{\frac{2E}{C}}\) \(V = \sqrt{\frac{2\times20}{100\times10^{-6}}} = \sqrt{20 \times 10^{5}}\)
02

Evaluate the voltage

Now plug in the values and calculate the voltage: \(V = \sqrt{20 \times 10^{5}} = 1000V\) So, the initial voltage of the capacitor is 1000V.
03

Calculate the initial charge

To find the initial charge on the capacitor, use the relationship: \(Q = CV\) where \(Q\) is the charge, \(C\) is the capacitance, and \(V\) is the initial voltage. With \(C = 100.0 \times 10^{-6}F\) and \(V = 1000V\), we have: \(Q = 100.0 \times 10^{-6}F \times 1000V = 100.0 \times 10^{-3}C\) Thus, the initial charge on the capacitor is \(100.0 \times 10^{-3}C\).
04

Calculate the time constant and current

To find the resistance of the lamp, first calculate the time constant, \(\tau\): \(\tau = RC\) We are given that the current reaches 5% of its original value in \(2.0ms\). That means the time constant \(\tau\) is \(2 \times 10^{-3}s\). We can then find the initial current, \(I_0\), using: \(I_0 = \frac{V}{R}\), where \(V = 1000V\) and \(R\) is the resistance we want to find. Using the given information, \(I(t) = I_0 e^{-t/\tau}\), where \(I(t)\) is the current at time \(t\), and \(I_0\) is the initial current. At \(t = 2.0 ms\) or \(2 \times 10^{-3}s\), the current reaches 5% of its original value, so \(I(t) = 0.05I_0\) That means, \(0.05I_0 = I_0 e^{-2\times10^{-3}/RC}\)
05

Calculate the resistance

Now, we can cancel out \(I_0\) to solve for the resistance \(R\): \(0.05 = e^{-2\times10^{-3}/RC}\) To find \(R\), take the natural logarithm of both sides and rearrange the terms: \(R = \frac{-2\times10^{-3}}{C \times \ln(0.05)}\) Plug in the known values and calculate \(R\): \(R = \frac{-2\times10^{-3}s}{100.0 \times 10^{-6}F \times \ln(0.05)} \approx 190.5\Omega\) So, the approximate resistance of the lamp is \(190.5 \Omega\).

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Most popular questions from this chapter

A parallel plate capacitor is constructed from two square conducting plates of length \(L=0.10 \mathrm{m}\) on a side. There is air between the plates, which are separated by a distance \(d=89 \mu \mathrm{m} .\) The capacitor is connected to a \(10.0-\mathrm{V}\) battery. (a) After the capacitor is fully charged, what is the charge on the upper plate? (b) The battery is disconnected from the plates and the capacitor is discharged through a resistor \(R=0.100 \mathrm{M} \Omega .\) Sketch the current through the resistor as a function of time \(t(t=0\) corresponds to the time when \(R\) is connected to the capacitor). (c) How much energy is dissipated in \(R\) over the whole discharging process?
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Show that \(A^{2} \times \Omega=W\) (amperes squared times ohms \(=\) watts).
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During a "brownout," which occurs when the power companies cannot keep up with high demand, the voltage of the household circuits drops below its normal $120 \mathrm{V} .\( (a) If the voltage drops to \)108 \mathrm{V},$ what would be the power consumed by a "100-W" light-bulb (that is, a light-bulb that consumes \(100.0 \mathrm{W}\) when connected to \(120 \mathrm{V}\) )? Ignore (for now) changes in the resistance of the light-bulb filament. (b) More realistically, the light-bulb filament will not be as hot as usual during the brownout. Does this make the power drop more or less than that you calculated in part (a)? Explain.
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