And that is the expression for the current vs. time curve for this circuit.

[TMac92]

Homework Statement

In the circuit below, the capacitor C is uncharged at time t=0, at which moment switch S is turned on. Current I is detected with the ammeter shown. Given the displayed paramters of the circuit elements, determine the temporal variation I(t) of the current from the moment the switch is closed.

(Image of circuit attached)

Homework Equations

I(t)=(emf/R) * e ^(-t/(RC))

The Attempt at a Solution

I believe I just need to plug all the variables into the equation above.
C=C
emf = emf1 (as noted in circuit diagram)
R=?

I was having trouble determining the total resistance R.
I understand there are two parallel branches of the circuit. If it one through one branch the resistance would be just R1, but if it went through the other branch, the resistance would be R1+R2. So would the total resistance be 1/R_total = (1/R1) + (1/(R1+R2))?

 

[gneill]

A complete solution to wrap up this thread:

When the switch is closed resistor $R_2$ and the capacitor $C$ are in parallel. We can re-draw the circuit, swapping these two branches to make the voltage divider clear:

When the switch is first closed the capacitor is uncharged and looks like a short circuit for the first instant. That means $R_2$ is bypassed and the full potential of the battery will be across $R_1$, yielding an initial battery current:

$I_o = \frac{E}{R_1}$

Eventually the capacitor will have charged to its final value and its current $I_c$ will be zero. Effectively it then looks like an open circuit to the rest of the circuit, leaving just the voltage divider comprised of $R_1$ and $R_2$. The battery just sees the two resistors in series, so the final current will be:

$I_f = \frac{E}{R_1 + R_2}$

So the source current will start at $I_o$ at time t = 0 and drop to at $I_f$ as t → ∞.

Since its a first order circuit the transition will follow an exponential curve as shown. What is required to pin down the details is the time constant, $\tau$.

In the circuit diagram above we can imagine replacing the battery and resistor divider with a Thevenin equivalent. Then we will have a simple RC circuit whose time constant will be $\tau = R_th C$. We don't need the Thevenin voltage here, just the Thevenin resistance.

Opening the circuit at a-b we can see that the Thevenin resistance is given by:

$R_th = R_1 || R_2 = \frac{R_1 R_2}{R_1 + R_2}$

So the circuit's time constant will be

$\tau = \frac{R_1 R_2}{R_1 + R_2} C$

We can now write the expression for the current by inspection of the current vs time sketch:

$I(t) = I_f + (I_o - I_f) e^{- \frac{t}{\tau}}$

We can plug in the various expressions for the parameters from their derivations above and rearrange for a "pleasing" form for the equation. Perhaps:
$$ I(t) = \frac{E}{R_1 + R_2} \left( 1 + \frac{R_2}{R_1} e^{- t\frac{R_1 + R_2}{R_1 R_2 C}} \right) $$