Question

let the capacitor be initially uncharged, and close the switch to position a at time t = 0. The battery begins charging the capacitor. Find the charge q(t) on the capacitor for any time , and the current I(t). Do the answers make sense for t = 0 and for t going to infinity?

by conservation of energy (per unit charge) we have

Voltage put into circuit = voltage used by circuit

Vo = IR + q/C

= R dq/dt + q/C

Separate variables and integrate (you get a natural log), use the initial condition q(0) = 0, and when you invert (to get an exponential), to find q(t), then you can find I = dq/dt from that.

When the switch is set to "b", the capacitor already has the charge q(t = infinity) = CVo, so when the switch is closed current starts flowing, where the charges on the plus plate run around to neutralize the charges on the minus plate.

Then q/C = IR where I = dq/dt, and integrate similar to how you did in the first part.

Answer 1

a)

let the capacitor be initially uncharged, and close the switch to position a at time t = 0. The battery begins charging the capacitor.

Using conservation of energy ( or Kirchhoff voltage law) we have

or

now, separating the variables and the integrating both sides we get:

here ln(k) is constant of integration.

now taking exponential both sides we get:

or

Now from initial condition at time t=0 capacitor is uncharged (i.e., q(t=0)=0)

hence, from eq(3)

or

substituting value of k in eq(3) we get:

Now, Using eq(5) at t=0, the charge stored in capacitor is given by:

which make sence, since at t=0 the capacitor is discharged.

Again, using eq(5) as t tends to infinite charge in the capacior is given by:

which implies that at t tends to infinity the capacitor charges upto its fulll capacity(CV0).

Now, the current in the circuit is givej by diffrentiating eq(4) with respect to time:

b)

When the switch is set to "b", the capacitor already has the charge

q(t = infinity) = CVo, so when the switch is closed current starts flowing, where the charges on the plus plate run around to neutralize the charges on the minus plate.

Then, using kirchoff's voltage law

where

(negative sign is because the charge is decreasing with time.)

Hence, we can write:

separating, the variables and integrating both sides we get

here ln(k') is constant of integration.

taking exponential both sides we get:

now at t=0 when switch is set to b the capcitor is charged fully

[i.e., q(t=0)=CV0]

Hence, using eq(9)

substituting k' in eq(9) we get:

as t tends to infinity the capacitor again discharge. which make sense.

and current is given by: