Resistors in parallel or series?

ErniePicks

Hey finding the thevenin equivalent circuit and am wondering if the 10 and 15 ohm resistors are in paralllel or series?

Reply 1

Squishy®

Series as they aren't parallel to each other

Reply 2

EricPiphany

Original post by Squishy®

Series as they aren't parallel to each other

Lol good one.

Reply 3

atsruser

Original post by ErniePicks

Hey finding the thevenin equivalent circuit and am wondering if the 10 and 15 ohm resistors are in paralllel or series?

In the Thevenin equivalent circuit, you will have replaced the voltage source by a short circuit. So the 20 ohm resistor will be in parallel with a 0 ohm resistor, giving a resistance of 0 ohms, leaving the 10 ohm and 15 ohm resistors in parallel across the load terminals.

Reply 4

ErniePicks

Original post by atsruser

What about when i'm calculating the open circuit voltage? they are still in parallel then as well? I'm not sure how i would calculate the open circuit voltage. Would i just use the voltage divider rule and do (15k/25k)*12V?

Also are you sure they are in parallel? The other guy said they are in series, i thought they were in parallel as well though. It's just i'm using the voltage divider rule on the calculation i used above for two resistors in parallel, i just thought i could use it because all the current would flow through the 15k resistor.

(edited 8 years ago)

Reply 5

transformersone

The 15 and 20 are quite clearly in parallel due to the way the power supply splits them up.

The 10 and 15 are not in parallel because if you cut out the 20 for example, it becomes a series.

Therefore, if you were to calculate the entire circuit resistance, the 10 and 15 effectively become one resistor.

Reply 6

uberteknik

Original post by ErniePicks

Hey finding the thevenin equivalent circuit and am wondering if the 10 and 15 ohm resistors are in paralllel or series?

I've just noticed this question and realised it's why you are having difficulty with series and parallel circuits in another thread/question you asked.

Serious advice: Forget Thevenin for the time being (come back to it later). It is a different concept to straight forward series and parallel resistances. Concentrate on understanding series and parallel combinations first and only then come back and ask about Thevenin. Otherwise it will confuse the hell out of you.

Reply 7

ErniePicks

Original post by uberteknik

lol it seems like people in this thread are confused as well, some people are saying the 10 and 15 are in series whilst others are saying it's parallel.

it's having the wire in between the 15 and 10 ohm resistor leading to the open circuit which is confusing me. Current would flow through the 10 ohm and then due to the open circuit the rest would flow through the 15 which would make me think it was in series even though it would be in parallel with the load.

Reply 8

atsruser

Original post by transformersone

The 15 and 20 are quite clearly in parallel due to the way the power supply splits them up.

The 10 and 15 are not in parallel because if you cut out the 20 for example, it becomes a series.

He wants to calculate the Thevenin equivalents, not to analyse the circuit as it stands.

Regardless it isn't correct to say, even for the circuit as shown, that the 15 and 20 are in parallel - as it stands, we have 15 || (10 + 20), if we ignore the voltage source.

[edit: I'm going to modify this a bit:

From the POV of calculating the Thevenin equivalent voltage, then we need to calculate the resistance as 10 || (20 + 15) but we need to find

V_{\text{th}}

across the 15 ohm resistor - however, we still don't have 15 || 20 as a single resistance in the circuit ]

(edited 8 years ago)

Reply 9

atsruser

Original post by ErniePicks

What open circuit are you thinking of? To calculate a Thevenin equivalent, you replace voltage sources with *short* circuits.

Reply 10

uberteknik

Original post by ErniePicks

Thévenin theory allows us to simplify the most complex circuit to a simple 'equivalent circuit' comprising a single voltage source and a single series resistance connected to a load.

It's use becomes apparent when there are several voltage sources in the circuit, like several batteries connected in various series and parallel combinations.

To reduce a circuit using Thévenin:

1) Label the load nodes A and B.

2) replace the load with an open circuit and calculate the potential difference across that (infinite resistance) open-circuit. (Use ohms law series and parallel combinations to accomplish). This becomes V_TH. (in other words V_AB)

3) replace the power sources with a short circuit and calculate the series/parallel equivalent resistance looking back in across the AB nodes. (once again the load is not connected). This becomes R_TH.

4) The Thévenin equivalent circuit connected to the load is then the calculated V_THas the supply and R_TH as the resistance in series with the load.

In essence, the load 'sees' the circuit connected to it, as if it were a battery with an internal resistance.

Wiki has a good explanation:

https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theoremtion:

(edited 8 years ago)

Reply 11

ErniePicks

Original post by atsruser

What open circuit are you thinking of? To calculate a Thevenin equivalent, you replace voltage sources with *short* circuits.

Between nodes A and B, it's an open circuit and has an open circuit voltage right?

Reply 12

atsruser

Original post by ErniePicks

Between nodes A and B, it's an open circuit and has an open circuit voltage right?

Assuming that nodes A and B are the two unlabelled terminals, then yes, that is right.

However, if you want to calculate the Thevenin equivalent resistance, you replace all voltage sources with short circuits, then you calculate the resistance of that new circuit between the terminals. If you do that you will find a total resistance between the terminals of 15 || 20 ohms.

Maybe I didn't make that clear earlier.

To calculate the Thevenin equivalent voltage, you just analyse the circuit as it stands, with an open source load. With your circuit, you will have a 10/15 ohm voltage divider in parallel with 10 ohms, so you will have 12 V across 20 || (10 + 15), then you need to find the voltage across the 15 ohm resistance.

Reply 13

uberteknik

Original post by atsruser

Replacing the 12v source with a short circuit and looking back at the nodes AB will produce:

15||10 ohms = 6 ohms

Original post by atsruser

To calculate the Thevenin equivalent voltage, you just analyse the circuit as it stands, with an open source load. With your circuit, you will have a 10/15 ohm voltage divider

Correct

Original post by atsruser

in parallel with 10 ohms, so you will have 12 V across 20 || (10 + 15), then you need to find the voltage across the 15 ohm resistance.

The 20 ohm resistor is across the battery. Current through the 20 ohms path is therefore independent of the 10 and 15 ohms voltage divider current path.

This is why the power source is replaced with a short circuit. Because anything directly in parallel with the supply (other than the node path) can be ignored.

The Thévenin equivalent voltage appearing at the AB nodes is therefore:

V_{AB} = V_S(\frac{15}{10 + 15}) = 7.2V

Making the Thévenin equivalent circuit replacing the original as:

V_{TH} = 7.2V

in series with

R_{TH} = 6\Omega

(edited 8 years ago)

Reply 14

atsruser

Original post by uberteknik

Replacing the 12v source with a short circuit and looking back at the nodes AB will produce:

15||10 ohms = 6 ohms

Right. My 20 should have been 10. That was a typo.

V_{AB} = V_S(\frac{15}{10 + 15}) = 7.2V

Making the Thévenin equivalent circuit replacing the original as:

V_{TH} = 7.2V

in series with

R_{TH} = 6\Omega

I agree with this, but I'm not sure if you are disagreeing with me.

All you seem to have done is to perform the voltage divider calculation that I omitted. And I don't understand why you've underlined part of your explanation. It's rather oddly worded - the current through the 20 ohms isn't really *independent* of the other path - it would change if we had 100 and 150 ohms instead. But otherwise it simply seems to repeat what I said (that we have 12 V across 20 || (10+15) ohms).

I guess that the point that you are trying to make is that the 20 ohm path is irrelevant to the calculation of

V_\text{th}

, which I possibly didn't make clear.

Reply 15

uberteknik

Original post by atsruser

Right. My 20 should have been 10. That was a typo.

I agree with this, but I'm not sure if you are disagreeing with me.

All you seem to have done is to perform the voltage divider calculation that I omitted. And I don't understand why you've underlined part of your explanation. It's rather oddly worded - the current through the 20 ohms isn't really *independent* of the other path - it would change if we had 100 and 150 ohms instead. But otherwise it simply seems to repeat what I said (that we have 12 V across 20 || (10+15) ohms).

I guess that the point that you are trying to make is that the 20 ohm path is irrelevant to the calculation of

V_\text{th}

, which I possibly didn't make clear.

Apologies for not making myself more clear, I'm not disagreeing with you. :smile:

The underlined part is an attempt to clarify to others (who are also reading this thread) why Thévenin methods work the way they do.

i.e. from a current path perspective, any resistance directly in parallel with the power source (not in the node current path) can be ignored.