Answered: An electrician discovers that he can't…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

An electrician discovers that he can't read the capacitance of one of his capacitors. He connects the unknown capacitor in series with another capacitor that has a capacitance of 12.1×10−6 F,12.1×10−6 F, and a battery that supplies a voltage difference of 361 V361 V across the capacitors. The unknown capacitor now has a charge of 0.00145 C,0.00145 C, what is its capacitance?

Expert Solution

Step 1

Given data:

The capacitance is $C = 12.1 \times 10^{- 6} F$
The voltage difference is $∆ V = 361 V$
The charge of the unknown capacitor is $Q_{u} = 0.00145 C$

The net voltage is equal to the sum of the voltage in the capacitors as:

$\begin{array}{rcl} ∆ V & = & V_{u} + V \\ ∆ V & = & \frac{Q_{u}}{C_{u}} + \frac{Q}{C} (Q = Q_{u}) \\ ∆ V & = & Q_{u} (\frac{1}{C_{u}} + \frac{1}{C}) \end{array}$

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