Sunday
April 20, 2014

Posts by Venkatarama


Total # Posts: 10

physics
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.?

physics
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 initial voltage of 1v and the other has no charge on it initially.What is the final voltages across the capacitor C2?

physics
can anybody answer this?

physics
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 initial voltage of 1v and the other has no charge on it initially.What is the final voltages across the capacitor C2?

Physics
The first answer is correct...final voltage will be: Qtotal/(C1+C2)=0.347

Physics
The first answer is correct...final voltage will be: Qtotal/(C1+C2)=0.347

Physics
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 conservation of charge: Qeff=(C...

Physics
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 since they are in paral...

physics
the answers would be: True: All the capacitors have the same charge, and the equivalent capacitance is less than the capacitance of any of the individual capacitors in the group. True:The largest potential difference appears across the capacitor with the smallest capacitance.

Physics
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

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