For the capacitor network shown in (figure 1) , the potential difference across ab is 68 v .

Answer
First we need the total equiv. cap.

C = 150•120/(150+120) = 67.7 nF

total charge stored Q = CV = 68v x 67.7 nF = 4533 nC or 4.533 µC

this charge is on each cap.

For C1 (left) V1 = Q/C = 4533 nC / 150nF = 30.22 volts

For C2 (right) V2 = Q/C = 4533 nC / 120nF = 37.78 volts

note that they add up to 68 volts as a check.

For C1, E1 = ½CV² = ½150nF•30.22² = 68490 nJ

or, rounding, 68 µJ

For C2, E2 = ½CV² = ½120nF•37.78² = 85640 nJ

or, rounding, 86 µJ

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