Three capacitors C1 = 1.5 µF, C2 = 9.5 µF, and C3 = 14.5 µF are connected as sho

Question

Three capacitors C1 = 1.5 µF, C2 = 9.5 µF, and C3 = 14.5 µF are connected as shown with a battery of voltage V = 18 V. (When entering units, use micro for the metric system prefix µ.)


(b) What is the charge on C1?


(c) What is the voltage across each capacitor?
V1 =
V2 =
V3 =

(d) What is the charge on C2 and C3?
Q2 =
Q3 =


http://www.webassign.net/ncsuplseem2/3-post-002.gif

Explanation / Answer

First find the equivalent capacitance of the circuit. C2 and C3 are in parallel, so they add to 9.5uF + 14.5uF = 24uF. In series with C1, the total capacitance is 1/(1/1.5uF + 1/24uF) = 1.412uF. (b) Charge Q = CV, and the charge is the same across series capacitors, so the charge across C1 is the equivalent circuit capacitance multiplied by the voltage source: 1.412uF * 18V = 25.412uC (c) voltage across capacitor V = Q/C --> we already know the charge and capacitance of C1, so V1 = 25.412uC/1.5uF = 16.94V. Since C2 and C3 are in parallel, the voltage across them is the same. Since the sum of all voltages in the loop must be 0, the voltages V2 and V3 both equal 18V - 16.94V = 1.06V. (d) The charge across C2 is Q2 = CV = 9.5uF * 1.06V = 10.059uC. The charge across C3 is Q3 = CV = 14.5uF * 1.07V = 15.353uC.

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