A 2.50- F capacitor is charged to 754 V and a 6.80- F capacitor is charged to 57

Question

A 2.50-F capacitor is charged to 754 V and a 6.80-F capacitor is charged to 574 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? [Hint: Charge is conserved.]

Determine the potential difference across the first capacitor.

Determine the potential difference across the second capacitor.

Determine the charge across the first capacitor.

Determine the charge across the second capacitor.

Explanation / Answer

Given

capacitors are  

let C1 = 2.50 F and v1 = 754 V

C2 = 6.80 F and v2 = 574 V

the initial charges of the capacitors are  

Q1 = C1*V1 = 2.50*10^-6*754 C = 0.001885 C

Q2 = C2*V2 = 6.8*10^-6*574 C = 0.0039032 C

the battery were disconnected and the capacitors are connected in parallel combination

so that the potential difference across each capacitor is same  

which is equal to

C1*V1+C2*V2 = V(C1+C2)

(2.50*10^-6*754+6.8*10^-6*574 ) = V(2.50*10^-6+6.8*10^-6)

V = 622.38 V

the potential acroos each capacitor is V1 = V2 = 622.38 V

and for the charge across each capacitor is  

from relation between charge , potential and capacitance

Q = C*V

Q1 = C1*V = 2.50*10^-6*622.38 C = 0.00155595 C = 155.595*10^-6 C

Q2 = C2*V = 6.8*10^-6*622.38 C = 0.004232184 C = 4232.184*10^-6 C

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