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

[itex]R_1 = R_2 = 4700 \Omega[/itex]
[itex]C = 0.060 F[/itex]
[itex]V = 12 V[/itex]S1 has been closed for a long time. At t = 0, S2 is closed. Sketch a graph of the currents [itex]I_1[/itex] through [itex]R_1[/itex] and [itex]I_2[/itex] through [itex]R_2[/itex].
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 [itex]R_2[/itex] is placed in parallel with it, there must be a 12 V potential difference across it, too. The current [itex]I_2[/itex] must be [itex]\frac{12}{4700}[/itex] then. The current through [itex]R_1[/itex] 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 [itex]\frac{12}{9400}[/itex].

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

[itex]R_1 = R_2 = 4700 \Omega[/itex]
[itex]C = 0.060 F[/itex]
[itex]V = 12 V[/itex]


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



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 [itex]R_2[/itex] is placed in parallel with it, there must be a 12 V potential difference across it, too. The current [itex]I_2[/itex] must be [itex]\frac{12}{4700}[/itex] then. The current through [itex]R_1[/itex] 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 [itex]\frac{12}{9400}[/itex].

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 [tex]C\frac{dV_C}{dt} = I_1 - I_2[/tex]
[tex]V_C - I_2R_2 = 0[/tex] and
[tex]I_1R_1 - V + I_2R_2 = 0[/tex]

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

After solving that system, you will find that :

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

and [tex]I_1 = \frac{V - I_2R_2}{R_1}[/tex]

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 [tex]V_C = \frac{VR_2}{R_1 + R_2}[/tex], which is also the final potential drop across the second resistor.

 

[kyle8921]

I would like to provide some additional insights and clarify some points in your explanation.

Firstly, your reasoning for the current through R_2 at t=0 is correct. The current through a capacitor at the moment of closing a switch is given by I = C*(dV/dt), where C is the capacitance and dV/dt is the rate of change of voltage. In this case, since the switch S1 has been closed for a long time, the voltage across the capacitor is constant (12V) and therefore the current through R_2 is simply 12V/4700Ω = 0.00255A.

Secondly, I would like to clarify that the current through R_1 at t=0 is not exactly 0, but it is very small. This is because the capacitor acts as an open circuit at the moment of closing the switch, allowing only a small amount of current to flow through R_1. This small current is given by the formula I = V/R, where V is the voltage across the capacitor (12V) and R is the resistance of R_1 (4700Ω). This results in a very small current of 0.00255A, which is negligible compared to the current through R_2.

Now, let's look at what happens in between t=0 and t=∞. As you correctly pointed out, at a long time after S2 is closed, the resistors are effectively in series and the current through each of them must be the same. However, the current does not suddenly jump from 0.00255A to 0.00128A (12V/9400Ω) at t=∞. Instead, the current gradually increases from 0.00255A to 0.00128A as the capacitor discharges. This discharge is exponential in nature and is given by the formula I = I_0*e^(-t/RC), where I_0 is the initial current (0.00255A), t is the time, R is the total resistance (9400Ω) and C is the capacitance (0.060F). This means that the current through R_2 will gradually decrease while the current through R_1 will gradually increase until they both reach the steady state value of 0.00128A at t=∞.

I hope this explanation helps