A circuit with three capacitors is shown below. The top capacitor, of unknown ca

Question

A circuit with three capacitors is shown below. The top capacitor, of unknown capacitance CX, has a charge of 36pC. The capacitor at bottom left, of 12pF, has a charge of 72pC. The voltage VB of the battery is unknown. The next 5 questions refer to this figure.

The charge on the 24pF capacitor at the bottom right is: _______

The value of the capacitance CX is: _______

The voltage VB of the battery is: _______

The equivalent capacitance between points A and B is: _______

The total energy stored in all the capacitors is: _______

Explanation / Answer

Since capacitors 12pF and 24 pF are in series combination they will have same change,

Thus q12=q24 = 72pC

V12= q12/C12 = 72pC/12pF = 6V

V24= q24/C24 = 72pC/24pF = 3V

V12+V24 = V12+24= 9V

Since capacitors 12pF + 24 pF and Cx are in Parallel combination they will have same voltage

Thus

VCx = V12+24 = 9V

CCx = qCx/VCx = 36pC/9V = 4pF

VB = VCX = V12+24=9V

Since capacitors 12pF and 24 pF are in series combination , their equivalent capacitance is addition of both

Thus,

Since C12 and C24 are in series combination

1/C1224 = 1/C12 +1/ C24 = 1/12pF +1/24 pF = 8 pF

Since C1224 and CCX are in parallel combination

The equivalent capacitance between points A and B = CTotal = CCX + C1224 = 4F+ 8pF =12pF

E =1/2CTotalVB^2 = ½*12pF*9^2 = 4.86*10^-10 J

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