Equivalent resistance with a capacitor

[RED119]

Homework Statement

So I need to find the equivalent resistance for the following RC circuit to calculate the time constant, but I got stuck in terms of finding out the equivalent resistance.

Homework Equations

V = IR

The Attempt at a Solution

So the thing I am not sure on is how do you calculate equivalent resistance when there is a capacitor in there? it counts as an element right? so the resistors are in neither series or parallel? or is there a way to calculate the equivalent resistance and I am just missing it? cause the resistors don't share two common but separate nodes with the capacitor in there

 

[gneill]

The basic procedure is to remove the capacitor from the circuit, then find the Thevenin Equivalent of the remaining network as seen from the terminals where the capacitor was formerly connected. The resistance that the capacitor "sees" is the Thevenin resistance.

If you haven't done Thevenin Equivalents yet, the essential idea is to replace the voltage source with a short circuit (an open circuit if it were a current source instead) and find the equivalent resistance at the terminals where the capacitor was connected.

 

[RED119]

The basic procedure is to remove the capacitor from the circuit, then find the Thevenin Equivalent of the remaining network as seen from the terminals where the capacitor was formerly connected. The resistance that the capacitor "sees" is the Thevenin resistance.

If you haven't done Thevenin Equivalents yet, the essential idea is to replace the voltage source with a short circuit (an open circuit if it were a current source instead) and find the equivalent resistance at the terminals where the capacitor was connected.

So do something like one of these? I don't totally understand how that will make it so you can find an equivalent resistance

So i think you are saying take out the capacitor in the second drawring, and have the two resistors be connected and then they are in parallel and that's the thevenin resistance?

 

[gneill]

Replace the voltage source with a short circuit (a piece of wire).

 

[RED119]

Replace the voltage source with a short circuit (a piece of wire).

so that would make the 10k in this example shorted?

 

[gneill]

so that would make the 10k in this example shorted?

It would indeed.

 

[RED119]

It would indeed.

so the equivalent resistance is just 50k? Why is the 10k shorted, or at least the capacitor sees is that way? logically speaking it seems like it should dissipate power from the battery with its placement, and have a voltage drop across it...

 

[gneill]

so the equivalent resistance is just 50k? Why is the 10k shorted, or at least the capacitor sees is that way? logically speaking it seems like it should dissipate power from the battery with its placement, and have a voltage drop across it...

Yes, but the capacitor never "sees" that resistor. The voltage source being fixed, creates an impenetrable wall beyond which no voltage changes in the rest of the circuit can pass. The battery can supply all the current that that resistor will use.

 

[RED119]

Yes, but the capacitor never "sees" that resistor. The voltage source being fixed, creates an impenetrable wall beyond which no voltage changes in the rest of the circuit can pass. The battery can supply all the current that that resistor will use.

that makes sense, so its almost like a sub-circuit that has no affect on the capacitor. Thanks so much for the help and explanations

 

[gneill]

Yup. If you wish to picture it another way, the voltage source can be duplicated and produces the same potential for both "sides" of the circuit:

 

[DaveE]

Are you sure you drew the schematic correctly? It's an unusual question because the 10K resistor is meaningless in the analysis, as others have described.

 

[Tom.G]

Are you sure you drew the schematic correctly? It's an unusual question because the 10K resistor is meaningless in the analysis, as others have described.

I think that was one of the points the question was trying to get across.
Another point was the concept of an ideal voltage source, having zero internal resistance, could consequently be replaced with a short to find the equivalent, Thevenin, circuit.