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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

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