# Quick question on calculating time constant RC?

1. Oct 30, 2012

### letsgo

1. The problem statement, all variables and given/known data

If you have this RC Circuit (ignore the yellow path of current):

http://i.imgur.com/Mi2bj.png

When you are calculating time constant=RC, what would you use for R?
R1=100 ohms, or Req=300 ohms

2. Relevant equations

time constant = RC

3. The attempt at a solution

I would think its just the 100 ohms, because initially, the current is only flowing through that resistor.

2. Oct 30, 2012

### Staff: Mentor

The 200 Ω resistor is in parallel with the capacitor, so you can swap their positions without changing the circuit in any way. If you do that, consider the equivalent resistance that the capacitor "sees" from the network feeding it (Think Thevenin equivalent).

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3. Oct 30, 2012

### letsgo

4. Oct 30, 2012

### Staff: Mentor

Sorry, that's not how it works here. We can't just provide answers to homework questions, we can only help you to work them out yourself via hints, corrections, or suggestions.

I can tell you that it's neither 100 nor 300 ohms...

5. Oct 31, 2012

### letsgo

Okay, so its 75 ohms, because you treated them as parallel. But why did you treat them as parallel?

6. Oct 31, 2012

### Staff: Mentor

Check your math: 100||200 is not 75.

They are treated as being in parallel because that is the equivalent resistance that the "source network" presents to the capacitor.

The usual approach to finding this value is to find the Thevenin equivalent of the network that the capacitor is connected to. However, in this case a thought experiment might suffice. Imagine that the circuit is allowed to settle over time and that the capacitor has reached its full charge. Now, at some time we'll call t = 0, the battery is replaced by a short circuit (effectively, the voltage source becomes 0V). Nothing else in the circuit involving capacitors or resistors has changed so the time constant should be the same as before. What's the time constant for this case? Can you now tell that the resistors are in parallel?

7. Oct 31, 2012

### letsgo

But no, I still don't see it. Because the way I see it, is if you're not considering the capacitor, then the resistors ARE in series. Because the current goes from the source to 200 to 100. There is no junction. See but this is when the circuit is in a steady state. Otherwise, immediately after, the capacitor would work as a wire and the current would bypass the 200 resistor, making only the 100 resistor count. So what I want to know, is when you're finding the time constant, do you take the resistance immediately after or when its in a steady state? I'm confused.

8. Oct 31, 2012

### Staff: Mentor

No, the capacitor is still there, it's the battery that you replace with a short (wire). The capacitor then "sees" two resistors in parallel.

Also, the 200 Ω resistor always "counts" because immediately after current begins to flow (no matter how small the time interval), any charge on the capacitor will place a potential across the 200 Ω resistor, which must then carry some current.

9. Oct 31, 2012

### letsgo

I finally get it.

Thank you! So this method of replacing the battery with a wire and remaking the circuit (the image you supplied me with), is this what finding the Thevinin equivalent is (I haven't learned this yet), and is this something that you have to do to solve this problem?

10. Nov 1, 2012

### Staff: Mentor

Any network of resistors and power supplies (voltages and/or currents) that provides some output at a pair of terminals can be replaced with an equivalent circuit model consisting of a voltage source in series with a single resistance. This is called its Thevenin Equivalent. In this case the "pair of terminals" happens to be where the capacitor connects, and we're concerned with the equivalent resistance of the network (Thevenin Resistance) so we can use it to determine the RC time constant.

The procedure for finding the Thevenin model for a network consists of two parts. The first is to find the open-circuit voltage at the output terminals (remove and load -- in this case remove the capacitor and find the voltage at the open terminals). This is called the Thevenin Voltage. The second step is to "suppress" any sources by replacing any voltage supplies with shorts and removing any current supplies, then finding the equivalent resistance at the output terminals. This gives the Thevenin Resistance.

The isn't the only possible approach to the problem. An alternative would be to write Kirchhoff's Laws equations and solve the resulting differential equations for the capacitor current or voltage. The time constant would pop out of that. Your choice