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?

Mar 24, 2017

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:

${V}_{B} - i R - \frac{Q}{C} = 0$

We can differentiate that wrt time:

$\implies 0 - \frac{\mathrm{di}}{\mathrm{dt}} R - \frac{i}{C} = 0$, noting that $i = \frac{\mathrm{dQ}}{\mathrm{dt}}$

This separates and solves, with IV $i \left(0\right) = \frac{{V}_{B}}{R}$, as:

${\int}_{\frac{{V}_{B}}{R}}^{i \left(t\right)} \frac{1}{i} \frac{\mathrm{di}}{\mathrm{dt}} \setminus \mathrm{dt} = - \frac{1}{R C} {\int}_{0}^{t} \setminus \mathrm{dt}$

$i = \frac{{V}_{B}}{R} {e}^{- \frac{1}{R C} t}$, 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:

${V}_{R} - {V}_{C} = 0 \implies i R = \frac{Q}{C}$

Note that, in the discharge process: $i = \textcolor{red}{-} \frac{\mathrm{dQ}}{\mathrm{dt}}$

Again we can differentiate that wrt time:

$\implies \frac{\mathrm{di}}{\mathrm{dt}} R = - \frac{i}{C}$

This separates and solves as:

${\int}_{i \left(0\right)}^{i \left(t\right)} \frac{1}{i} \frac{\mathrm{di}}{\mathrm{dt}} \setminus \mathrm{dt} = - \frac{1}{R C} {\int}_{0}^{t} \setminus \mathrm{dt}$

$\implies i = i \left(0\right) {e}^{- \frac{t}{R C}}$

In this instance, because the capacitor is fully charged and so has voltage ${V}_{B}$, we know that $i \left(0\right) = {V}_{B} / R = \frac{12}{20} = 0.6 A$.

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

And so:

$i \left(t\right) = 0.6 {e}^{- \frac{t}{R C}}$

Finally at $t = 3$ we have :

$i \left(3\right) = 0.6 {e}^{- \frac{3}{20 \cdot {10}^{- 2}}} = 1.8 \times {10}^{- 7} A$