Solving Circuit Problem: Currents I_1 & I_2 at t=0 & Long Time

[mooshasta]

I came across the following problem recently:

http://www.tubaroo.com/stuff/circuit.gif

R_1 = R_2 = 4700 \Omega
C = 0.060 F
V = 12 VS1 has been closed for a long time. At t = 0, S2 is closed. Sketch a graph of the currents I_1 through R_1 and I_2 through R_2.
I wasn't really sure where to begin with this. I ended up with the following graph:

http://www.tubaroo.com/stuff/igraph.gif

I reasoned that at t=0, the entire voltage (12 V) appears across the capacitor (since S1 had been closed for a long time), so when R_2 is placed in parallel with it, there must be a 12 V potential difference across it, too. The current I_2 must be \frac{12}{4700} then. The current through R_1 is 0 at t = 0 because S1 had been closed for a long time. At a long time after S2 is closed, however, the resistors are effectively in series with one another. Therefore the currents through each must be the same, and equal to the voltage of the battery over the sum of the resistances, or \frac{12}{9400}.

I can't figure how exactly the graphs would look between the two points, but this is my best guess.Can anyone help explain to me exactly what happens in between? Much appriciated.

 

[Curious3141]

I came across the following problem recently:

http://www.tubaroo.com/stuff/circuit.gif

R_1 = R_2 = 4700 \Omega
C = 0.060 F
V = 12 V


S1 has been closed for a long time. At t = 0, S2 is closed. Sketch a graph of the currents I_1 through R_1 and I_2 through R_2.



I wasn't really sure where to begin with this. I ended up with the following graph:

http://www.tubaroo.com/stuff/igraph.gif

I reasoned that at t=0, the entire voltage (12 V) appears across the capacitor (since S1 had been closed for a long time), so when R_2 is placed in parallel with it, there must be a 12 V potential difference across it, too. The current I_2 must be \frac{12}{4700} then. The current through R_1 is 0 at t = 0 because S1 had been closed for a long time. At a long time after S2 is closed, however, the resistors are effectively in series with one another. Therefore the currents through each must be the same, and equal to the voltage of the battery over the sum of the resistances, or \frac{12}{9400}.

I can't figure how exactly the graphs would look between the two points, but this is my best guess.


Can anyone help explain to me exactly what happens in between? Much appriciated.

Your answer is essentially correct, and the shape of your graph is the correct exponential one.

To appreciate what is actually going on, you need to set up a system of equations involving Kirchoff's laws, and solve them (this will involve solving a first order differential equation).

The equations are C\frac{dV_C}{dt} = I_1 - I_2
V_C - I_2R_2 = 0 and
I_1R_1 - V + I_2R_2 = 0

which are basically Kirchoff's first and second laws applied to the loops and junctions of this circuit. V_C is the potential difference across the capacitor.

After solving that system, you will find that :

I_2 = \frac{V}{R_1 + R_2}(1 + \frac{R_1}{R_2}e^{-\frac{R_1 + R_2}{CR_1R_2}t})

and I_1 = \frac{V - I_2R_2}{R_1}

which will correspond to your graph.

In simple terms, you can visualise the capacitor as starting out with a charge of V volts, then starting to discharge across R2 when S2 is thrown. The discharge current will be exponentially decreasing with time. It will not discharge completely, instead, the system reaches steady state when the voltage across the capacitor is V_C = \frac{VR_2}{R_1 + R_2}, which is also the final potential drop across the second resistor.