GREPhysics.NET: GR0177 #10

GR0177 #10

Problem

GREPhysics.NET Official Solution

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Electromagnetism $\Rightarrow$ }Capacitors

The problem gives $C_1=3\mu F$ and $C_2=6 \mu F$ and V=300 V.

The capacitors are connected in series, and thus $C_{eq}=C_1C_2/(C_1+C_2)=18/9=2\mu F$ . (Recall that series and parallel capacitor formulae are opposite to that of resistors.)

Energy is $1/2 C_{eq} V^2 =9E4 \times 1E-6 = 9E-2$ , as in choice (A).

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Comments

walczyk
2011-04-06 23:16:56

off the subject but how do you find the voltages over the two capacitors?

busy3
2011-11-08 14:14:54

V1=Q1/C1 , V2=Q2/C2, we know the C's and Q1=Q2 (in series). We can eliminate Q from the equations and get a ratio. We also know V1+V2 =300

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h0dgey84bc
2007-09-21 05:14:03

if C=2E-3 F, and V^2=9E4, then surely 0.5*C*V^2 is

=1E-3*9E4
=90J

not 0.09J, where does the 1E-6 come from in your answer, confused

h0dgey84bc
2007-09-21 05:15:35

sorry i read micro as milli, my apologies

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if C=2E-3 F, and V^2=9E4, then surely 0.5*C*V^2 is =1E-3*9E4 =90J not 0.09J, where does the 1E-6 come from in your answer, confused

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