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Unformatted text preview: 23. (a) and (b) We note that the charge on C3 is q3 = 12 µC – 8.0 µC = 4.0 µC. Since the charge on C4 is q4 = 8.0 µC, then the voltage across it is q4/C4 = 2.0 V. Consequently, the voltage V3 across C3 is 2.0 V ⇒ C3 = q3/V3 = 2.0 µF. Now C3 and C4 are in parallel and are thus equivalent to 6 µF capacitor which would then be in series with C2 ; thus, Eq 2520 leads to an equivalence of 2.0 µF which is to be thought of as being in series with the unknown C1 . We know that the total effective capacitance of the circuit (in the sense of what the battery “sees” when it is hooked up) is (12 µC)/Vbattery = 4µF/3. Using Eq 2520 again, we find 1 1 3 +C = ⇒ 2 µF 4 µF 1 C1 = 4.0 µF . ...
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This note was uploaded on 01/25/2011 for the course PHYS 221 taught by Professor Hk during the Spring '10 term at The Petroleum Institute.
 Spring '10
 HK
 Physics, Charge

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