Question

The circuit in the figure has been in position a for a long time, then the switch is thrown to position b. With `Vb = 12 V, C = 10 mF, R = 20 W`. a.) What is the current through the resistor before/after the switch? b) capacitor before/after c) at t=3sec?

Answer 1

See below

Explanation:

[NB check units of resistor in question, assume it should be in `Omega`'s]

With the switch in position a, as soon as the circuit is complete, we expect current to flow until such time as the capacitor is charged to the source's `V_B`.

During the charging process, we have from Kirchoff's loop rule:

`V_B - V_R - V_C = 0`, where `V_C` is the drop across the capacitor's plates,

Or:

`V_B - i R - Q/C = 0`

We can differentiate that wrt time:

`implies 0 - (di)/(dt) R - i/C = 0`, noting that `i = (dQ)/(dt)`

This separates and solves, with IV `i(0) = (V_B)/R`, as:

`int_( (V_B)/R)^(i(t)) 1/i (di)/(dt) \ dt = - 1/(RC) int_0^t \ dt`

`i = (V_B)/R e^(- 1/(RC) t)`, which is exponential decay .... the capacitor gradually charges so that the potential drop across its plates is equal to source `V_B`.

So, if the circuit has been closed at a for a long time, then `i = 0`. So no current through either the capacitor or resistor before the switch to b.

After the switch to b , we are looking at an RC circuit, with the capacitor discharging to the point the drop across its plates is zero.

During the discharging process, we have from Kirchoff's loop rule:

`V_R - V_C = 0 implies i R = Q/C`

Note that, in the discharge process: `i = color(red)(-) (dQ)/(dt)`

Again we can differentiate that wrt time:

`implies (di)/(dt) R = - i/C`

This separates and solves as:

`int_(i(0))^(i(t)) 1/i (di)/(dt) \ dt = - 1/(RC) int_0^t \ dt`

`implies i = i(0) e^(- t/(RC))`

In this instance, because the capacitor is fully charged and so has voltage `V_B`, we know that `i(0) = V_B/R = 12/20 = 0.6A`.

That is the current immediately the switch is closed at b.

And so:

`i(t) = 0.6 e^(- t/(RC))`

Finally at `t = 3` we have :

`i(3) = 0.6 e^(- 3/(20 cdot 10^(-2)) ) = 1.8 times 10^(-7) A`