Question

1. Five hundred 8.20-µF capacitors are connected in parallel and then charged to a potential of 23.0 kV. For how long will the stored energy light a 200.0-W bulb until no energy remains in the capacitors?  

( )s

2. A 9.50-V battery is connected across two capacitors,

C_A = 11.5 µF

and

C_B = 2.0 µF,

connected in series.

(a) What is the equivalent capacitance of the two capacitors?
( ) µF

(b) How much energy is stored by such an equivalent capacitor?
( )J

(c) What is the charge on each of the capacitors?

QA	=	( )C
QB	=	( )C

(d) How much energy is stored in each of the capacitors in this circuit?

UA	=	( )J
UB	=	( )J

Answer 1

1. For capacitors connected in parallel, their capacitance get added. So 500 capacitors each of capacitance 8.2 will have total capacitance

The energy stored will be

So the time till which it will power a 200 W bulb is

2.a. When capacitors connected in series their equivalent capacitance is givn as

2.b. Energy stored is given by

2.c. In series, charge on both the capacitors are same. Charge is given by

2.d. In series connectrion, the potential is different. The potentiall is given by

The energy stored will be

Energy stored in 1st capacitor is

Energy stored in 2nd capacitor is