Need help with part b) of the following question. Full explanation required for

Question

Need help with part b) of the following question. Full explanation required for points. Thanks!

A 25.0

Need help with part b) of the following question. Full explanation required for points. Thanks! A 25.0 mu F capacitor and a 40.0 mu F capacitor are charged by being connected across separate 65.0 V batteries. Determine the resulting charge on each capacitor. Need help with part b) of the following question. Full explanation required for points. Thanks! A 25.0 mu F capacitor and a 40.0 mu F capacitor are charged by being connected across separate 65.0 V batteries. Determine the resulting charge on each capacitor.

Explanation / Answer

a)

charge on the capacitor 40uF, Q1 = C1*V = 40*10^-6*65 = 2.6*10^-3 C

charge on capacitor 25uF , Q2 = Q2*V = 25*10-6*65 = 1.625*10^-3 C

b)

After connecting, the charge distribution is such that voltage difference between the two capacitors is same. At the same time, some charges recombine. and then redistribution of remaining charges happen in a manner such that the two capacitors are in parallel.

charge remaining after recombination, Qnet = Q1-Q2 = (2.6-1.625)*10^-3 = 9.75*10^-4 C

net capacitance after recombination,Cnet = C1+C2 = 25+40 = 65 uC <------ parallel connection

Let after connection, the voltage difference between both be V' (ie both have same potential difference).

Let charge on C1 be = Q1'

charge on C2 be Q2'

We know, net charge(Q) = net capacitance(C)*net voltage(V')

So, V' = Qnet/Cnet = 9.75*10^-4/(65*10^-6) = 15 V <-----------------answer (potential across both 25uF and 40 uF)

So, Q1' = C1*V' = 40*10^-6*15 = 6*10^-4 = 600 uC <------------charge on 40uF

Q2 = C2*V' = 25*10^-6*15 = 3.75*10^-4 C = 375 uC<-------------charge on 25 uF

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