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lets say two capacitors C1 and C2 1uF each are connected in series with a voltage source 5v. One of the capacitors C2 has an
1. 2 answers
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lets say two capacitors C1 and C2 1uF each are connected in series with a voltage source 5v. One of the capacitors C2 has an
1. 1 answer
2. 644 views

what about the initial 1v on capacitor c2. Since the source is at 5v ,this capacitor charge cannot get redistributed.Hence should it not add to the final value across c2? can somebody throw some light on this.?
can anybody answer this?
a.Ceq=8mf. b.16v since they are in parallel to the source. c.q=5*16 and q=3*16 respectively. d. 128mC is the total charge
Ceq(c1,c2):3uF. This combination is in series with the 3uF capacitor and both are connected to the 15v battery. The voltage will get divided equally since they are both of the same value(7.5v). Hence energy would be. Both c1 and c2 have 7.5v across them
The first answer is correct...final voltage will be: Qtotal/(C1+C2)=0.347
Depending on which way they are wired:assume they are wired positive plate to negative plate of each other. Then Qeff=16*4-7*4=36uC.(If wired positive plate to positive plate then charges will add up) This charge will get redistributed. Applying

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