(a) the equivalent capacitance of thecapacitors in the figure above 1 µF (b) the
ID: 1742582 • Letter: #
Question
(a) the equivalent capacitance of thecapacitors in the figure above1 µF
(b) the charge on each capacitor
on the right 34.20 µFcapacitor 2 µC on the left 34.20 µFcapacitor 3 µC on the 28.20 µF capacitor 4 µC on the 6.00 µF capacitor 5 µC
(c) the potential difference across each capacitor
on the right 34.20 µFcapacitor 6 V on the left 34.20 µFcapacitor 7 V on the 28.20 µF capacitor 8 V on the 6.00 µF capacitor 9 V on the right 34.20 µFcapacitor 2 µC on the left 34.20 µFcapacitor 3 µC on the 28.20 µF capacitor 4 µC on the 6.00 µF capacitor 5 µC
Explanation / Answer
The circuit is shown in the following figure: (a)The equivalent capacitance of the parallel capacitors 6.00F and C2 F is C = 6.00 + C2= 6.00 + 28.20 = 34.20 F The equivalent capacitance of the capacitors C1F,C and C1 F is (1/C') = (1/C1) + (1/C) + (1/C1) Here,C1= 34.20 F or (1/C') = (1/34.20) + (1/34.20) + (1/34.20) or (1/C') = (1 + 1 + 1/34.20) = (3/34.20) or C' = (34.20/3) F = 11.4 F (b)The charge on the capacitors C1 F,C andC1 F is same.This is because the capacitors areconnected in series and when the capacitors are connected in seriesthen the charge on the capacitors is same.Therefore,the charge onthe capacitors is Q = C' * V or Q = 11.4 * 10-6 * 9.00 = 102.6 * 10-6C = 102.6 C Therefore,the charge on the right and left 34.20 Fcapacitors are 102.6 C. Let the charge on the 28.20 F and 6.00 F capacitorbe Q1 and Q2.Therefore,the potentialdifference across the two capacitors is Q = C * V' or V' = (Q/C) or V' = (102.6 * 10-6/34.20 * 10-6) = 3V Therefore,the charge on the two capacitors is Q1= 28.20 * 10-6 * V' or Q1= 28.20 * 10-6 * 3 = 84.6 *10-6 C = 84.6 C and Q2= 6.00 * 10-6 * V' or Q2= 6.00 * 10-6 * 3 = 18 *10-6 C = 18 C (c)Let the potential difference across the capacitorsC1 F,C and C1 F beV1,V and V2.Therefore,we get Q = C1* V1 or V1= (Q/C1) = (102.6 *10-6/34.20 * 10-6) = 3 V, Q = C * V or V = (Q/C) = (102.6 * 10-6/34.20 *10-6) = 3 V and V2= (Q/C1) = (102.6 *10-6/34.20 * 10-6) = 3 V Therefore,the potential difference across the right and leftcapacitors is 3 V. The potential difference on the two capacitors 28.20 Fand 6.00 F capacitors is 3 V.This is because the capacitors28.20 F and 6.00 F are connected in parallel and thepotential difference on the capacitors when they are connected inparallel is same. or V1= (Q/C1) = (102.6 *10-6/34.20 * 10-6) = 3 V, Q = C * V or V = (Q/C) = (102.6 * 10-6/34.20 *10-6) = 3 V and V2= (Q/C1) = (102.6 *10-6/34.20 * 10-6) = 3 V Therefore,the potential difference across the right and leftcapacitors is 3 V. The potential difference on the two capacitors 28.20 Fand 6.00 F capacitors is 3 V.This is because the capacitors28.20 F and 6.00 F are connected in parallel and thepotential difference on the capacitors when they are connected inparallel is same.Related Questions
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