RC ciruit...pls help...please?

Question

Consider the electrical circuit in http://www.uploadimage.org/images/9b2zj6gf6x6bs259c0gz.gif

The internal resistance of the battery may be ignored. Initially, the switches are both open and the capacitor has no charge. Close switch S1 at a certain time. At a sufficiently long time after S1 was closed, the capacitor is fully charged and the circuit becomes steady.

(a)During the charging period, how much work is done by the battery?

(b)During the charging period, how much heat is emitted by R1?

(c)How much charge is stored in C?

(d)If both S1 and S2 were closed and enough time has elapsed, how much charge is stored in C?

(e)Just after S1 is closed, what is the current through R1?

pls help

Answer 1

For a capacitor,

Q = C * V

Where charge = Q, voltage = V capacitance = C

(c)   Q = 2*10^-6 * 6 = 1.2 * 10^-5 C

(a) It's a 6 v battery, so it will do 6 Joules of work for each Coulomb of charge that flows.

Work done = 1.2 * 10^-5 * 6 = 7.2 * 10^-5 Joules

(b)  Energy emitted by R1 = (work done by battery) - (energy stored in capacitor)

Energy stored in capacitor = 0.5 CV^2

E = 0.5 * 2 * 10^-6 *36 = 3.6 * 10^-5 J

R1 therefore emits (7.2-3.6) * 10^-5 J = 3.6 * 10^-5 J

(d) After sufficient time with both switches closed, R1 and R2 are a simple potential divider with 6 * 2/(2+4)  = 2 volts across C

Q = CV = 2 * 10^-6 * 2 = 4 micro Coulombs

(e) Just after S1 is closed, p.d. across C = 0, so pd across R1 = 6v

I = V/R = 6/4 = 1.5A