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Chapter 25: Problem 57

A capacitor is charged until it holds \(5.0 \mathrm{J}\) of energy, then connected across a \(10-\mathrm{k} \Omega\) resistor. In \(8.6 \mathrm{ms}\), the resistor dissipates 2.0 J. Find the capacitance.

Short Answer

Expert verified
The capacitance of the capacitor is approximately 0.6 Farads.

Step by step solution

01

Setup the initial energy condition

Start by acknowledging that the initial energy of the capacitor \(E_i\) is given as 5.0 J.
02

Compute the final energy

It's given that a certain amount of energy – 2.0 J – has been dissipated by the resistor. This energy should come from the capacitor, which therefore should have finally \(E_f = E_i - 2.0 = 5.0 - 2.0 = 3.0J\).
03

Use the formula for energy stored in capacitor

The energy stored in a capacitor can be given by \(E = 0.5 * C * V^2\), where C is the capacitance and V is the voltage. However, remember that energy should be the same before and after the resistor dissipated its energy. Therefore, we should set up a proportion based on the formula for electrical energy: \(0.5 * C_i * V_i^2 = 0.5 * C_f * V_f^2\). As energy was conserved, \(C_i\) and \(C_f\) should be the same, i.e., the capacitance of the capacitor did not change due to the energy dissipation. The equation is then simplified to \(V_i^2 = V_f^2\).
04

Solve for the capacitance

You solve \(V_i^2 = V_f^2\) for capacitance by using the reduced expression for energy \(E = 0.5 * C * V_f^2\). The expression for \(V_f^2\) (after rearranging) is: \(V_f^2 = 2*E_f/C\). Substituting this expression into our previous equation, we obtain the capacitance \(C\): \(2*E_i/C = 2*E_f/C\) which simplifies to \(C = E_f/E_i\). Given \(E_f = 3.0 J\) and \(E_i = 5.0 J\), we find \(C = 3/5 = 0.6 F\).

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