I came across the following problem recently:(adsbygoogle = window.adsbygoogle || []).push({});

http://www.tubaroo.com/stuff/circuit.gif [Broken]

[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 [Broken]

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.

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# Homework Help: RC circuit

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