# Finding the Equivalent Resistance of a Circuit

Discussion in 'Homework Help' started by wiz0r, May 5, 2008.

1. ### wiz0r Thread Starter Active Member

May 2, 2008
64
0
Hello, I need to find the equivalent resistance of the following circuit that can be found in the attachment. Also, I need to find if the C, L and Req are in series or parallel.

What I did was; since there's no current passing through R1, R2, and R3, I assume that Req = R4. Therefore, the capacitance, inductance and the equivalent resistance are in Series.

Am I right? Or is there something wrong?

Any help will be gladly appreciated!

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2. ### Caveman Senior Member

Apr 15, 2008
471
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You need to state what the resistance is across. My guess is that you are looking for resistance between the open points between R1 and R3, but you are not clear on this.

I also assume that you are looking for the equivalent DC resistance of the circuit, which means that the reactive components (inductor and capacitor) basically simplify out of the circuit.

First of all, you are wrong on your assumptions. Resistance exists whether or not current flows. It is a property not a state. A 100k resistor that is not in a circuit will still have a resistance of 100k.

Since I assume you are looking at the equivalent DC resistance, the first step is to remove the reactive components. So, C1 is an open circuit, effectively removing it from the circuit. And L1 is a short circuit, which removes it.

Now R4 is in series with R3, so calculate that.
R1 is in parallel with R2, so calculate that.
Then those two results are in series with each other, so calculate that.

3. ### wiz0r Thread Starter Active Member

May 2, 2008
64
0
Hello Caveman,

First of all thanks for the detailed responce, I'm really grateful. It really helped me understand quite a few things.

Now, a few things I want to make clear, because I wasn't clear enough on my first post.

I'm trying to solve an RLC circuit, that I will attach in this post. Obviously, what I need to do first is to know if the R, C and L are either in Series or in Parallel(haven't covered series-parallel RCL circuits on my class).

So, what I do is to remove the voltage source by a short and the current source by an open circuit, then assume the switch will be opened at t>0, so there will be an open circuit where the switch was.

After doing that, I got the circuit that I posted on my first post. So, I can't really do what you told me to do, not on this step anyway. Or, am I wrong? Please enlighten me.

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4. ### Caveman Senior Member

Apr 15, 2008
471
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You are still not being clear. What do you mean by "solve an RLC circuit"?
Are you saying that the circuit is completely stabilized, and the switch is opening at t=0. and you want to know what the voltages and currents some parts will be? Exactly, what do you need to know? Voltage on C1, current on C1, voltage on R4?

Two components (R,L,C or mixture) are in series if they have the same current through them. For example L1 and R4 are in series.

Two components are in parallel if they have the same voltage across them. For example R1 and R2 are in parallel.

5. ### wiz0r Thread Starter Active Member

May 2, 2008
64
0
I'm sorry, what the problem ask me is to find V(t) which is the voltage of the Resistance number 4.

Basically, find V(t) of R4.

I know the procedure of finding V(t), basically I need to find IL(t) AKA the current of the inductance in a function of time to be able to find the voltage of R4.

Im trying to find if R, L, and C are in series or parallel to determine weather the circuit is overdamped, underdamped or critical.

6. ### wiz0r Thread Starter Active Member

May 2, 2008
64
0
Caveman, sorry, here is the original circuit. It is on the attachment.

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7. ### Caveman Senior Member

Apr 15, 2008
471
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How to solve this problem:

1. Determine what the state of the circuit is before the switch closes. Let's assume that the circuit has reached steady-state, so everything has settled down. The capacitor is like an open-circuit and the inductor is now a short circuit. The current is 2A and all of it is going through R4, so all of it is going through L1. The voltage across R4 = 2 ohms * 2 amps = 4 volts. This is also the voltage across C1.

So just before t=0,
V(C1) = 4V. I(C1) = 0A
V(L1) = 0V. I(L1) = 2A
V(R1) = 4V. I(R1) = 2A

Now, here are some hints on how to simplify the circuit after the switch closes.
1. Find the norton equivalent of V1 and R1.
2. Then add the new current source to I1.
3. Then convert the resulting current source and the norton resistance to their equivalent thevenin equivalent.
4. At this point you have a voltage source in series with its thevenin resistance in series with R3, so you can add R3 and the thevenin resistance.

Here's the resultant circuit:

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8. ### wiz0r Thread Starter Active Member

May 2, 2008
64
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Wow, that looks great.

I'll get to it now, and post my doubts and stuff later.

Once again, thanks a lot Caveman, you are the man.

9. ### wiz0r Thread Starter Active Member

May 2, 2008
64
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Need to post this new circuit.

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10. ### wiz0r Thread Starter Active Member

May 2, 2008
64
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Adding another circuit for reference.

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11. ### Caveman Senior Member

Apr 15, 2008
471
1
Before t=0, the circuit is steady state with the switch closed, so it is simply R1 and R2 acting as a divider. So the state of circuit is:
V(C) = 37.5V, I(C) = 0A
V(L) = 0V, I(L) = 2.5A

At t=0 the circuit opens, the inductor will try to maintain the current and you have a series RLC circuit decaying.

12. ### wiz0r Thread Starter Active Member

May 2, 2008
64
0
Oh, there we go. That's what I needed.

Thanks a lot, Caveman. Wish you were my professor. Lol, I'll talk to you later. <3