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### Network # Potential divider problem

#### pyrohaz

Oct 28, 2012
33
Hey guys, I'm learning the rules on how to correctly debug a circuit and I have a problem as shown, its a potential divider with each resistor connected to a different voltage point. With the voltages and values given, how can I calculate the current through each resistor?

Thanks, Harris

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,615
There are many ways to attack this problem. Here's just one:
name the resistors and currents: Set up the node and loop equations:
i1+i2+i3 = 0 A (the sum of al currents in a node is equal to zero)
i1*R1-i2*R2=10V - (-5V) = 15V
i1*R1-i3*R3= 10 V
You now have three equations with three unknowns. Solve the system.

Harald

By the way: this looks like homework. If it is, why didn't you post it in the homework section?

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
There are probably lots of ways to solve this problem. The way that appeals to me is to use Thevenin equivalents.

I'll explain how to do it, without giving you the answer. (This looks like a classic homework problem; if so, you need to know HOW to solve it, not the solution.)

Two resistors in series, connected between two different voltage points, can be reduced to an equivalent consisting of one resistor connected to some intermediate voltage point. The voltage is calculated using the voltage divider equation, and the resistance is calculated using the parallel resistor equation.

You can start with any two resistors. Let's ignore R2 for the moment and calculate the Thevenin equivalent of R1 and R3. We have R1 (100) from +10V in series with R3 (300) to 0V. This forms a voltage divider; with nothing connected to the middle node, its voltage is +7.5V and its resistance is 75. Now you add R2, forming a new voltage divider with R1|R3 (75) from +7.5V in series with R2 (200) to -5V. Total voltage across R1|R3 and R2 is 12.5V across a resistance of 275, so the voltage across R2 is 9.09091V.

From this you can calculate the centre node voltage, and from that, the voltage across each resistor and the current flowing in each resistor.

Last edited:

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
You could also do it by superposition.

This cat can be skinned many ways

#### pyrohaz

Oct 28, 2012
33
Hey guys, it wasn't a homework program, i'm currently reading "Basic electronics for Scientists and Engineers" and there was a problem in it related to that but I put different values on the schematic, I want to learn the method vs just the answer so I appreciate the help there. I found the thevenin method much easier to comprehend than the first method (kirchoff's current and voltage laws if I remember correctly?) so thank you for all the replies!

In reply to KrisBlueNz, I understand how to work out the voltage across R2 (9.09v)

So would my method be:
R1||R3 = (1/100 + 1/300)^-1
Then 10/R1||R3 = 7.5v @75ohm (as you stated)

To then work out the voltage across the 200ohm resistor:
R1||R3 in voltage divider with the 200ohm resistor
7.5v to -5v is equivalent to 12.5v to 0v
12.5*200/(200+75) = 9.09v

Now i've calculated the voltage across R2, the voltage at the node that all the resistors meet should be 9.09v-5v = 4.09v
Therefore current through R3 = 4.09/300 = 13.6mA
and current through R1 will be (10-4.09)/100 = 59.1mA

Is this correct?
Thanks a lot guys, really appreciate it #### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Kirchoff's laws are the general case. It would be wise to try them out at least once.

The other methods are simplifications that work in certain subset cases.

I'll see if I can walk you through the superposition one later on (it can be even easier)

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Now i've calculated the voltage across R2, the voltage at the node that all the resistors meet should be 9.09v-5v = 4.09v
Therefore current through R3 = 4.09/300 = 13.6mA
and current through R1 will be (10-4.09)/100 = 59.1mA
Is this correct?
Yes, that's all correct. And the current through R2 is 45.5 mA.
The current through R1 feeds INTO the summing node (assuming conventional current) and the R2 and R3 currents feed OUT. So the total current is:
59.1 - 45.5 - 13.6 = 0 mA as a final check. Good work, you understand this stuff well.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
I'll see if I can walk you through the superposition one later on (it can be even easier)
I'd be interested to see that too, Steve.

#### pyrohaz

Oct 28, 2012
33
I'd be interested to see that too, Steve.
+1 to that, i'd really appreciate that Steve #### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
OK, for superposition you turn off all (independent) voltage and current sources, then turn them on one at a time, before adding the results together.

Turning off a current source means replacing it with an open circuit. Turning off a voltage source means replacing it with a short.

So in this example we have 2 Voltage sources, +10V and -5V.

(this becomes a lit easier to follow if you draw in the voltage sources and reference them to ground. Replacing them with a short when they're turned off will make the topology more obvious)

First, let's turn off the -5V source. Effectively this gives us R2 and R3 in parallel to ground. the total resistance is 100R in series with 120R. The current from the 10V source (i1) is therefore 10/(100+120) = 45.45mA. Thus the voltage at junction of the three resistors is 5.455V. The current through R2 (i2) is -27.27 mA and through R3 (i3) is -18.18 mA

Next turn off the 10V source. Effectively this now gives us R1 and R3 in parallel, in series with R2. The total resistance is thus 275 ohms. Thus i2 is -18.18 mA and the voltage at the junction of the resistors is -1.364V. This gives us i1 as 13.64mA and i3 as 4.55mA

Now we can add them all up

i1 = 45.45 + 13.64 = 59.09 mA
i2 = -27.27 + -18.18 = -45.45 mA
i3 = -18.18 + 4.55 = -13.63 mA
Vx = 5.455 + -1.364 = 4.091 V

Some quick tests i1 + i2 + i3 should be zero (and it's 0.01 which is due to rounding)

Vx using R3 and i3 gives us 4.089 V
Vx using R1 and i1 gives us 4.091 V
Vx using R2 and i2 gives us 4.090 V

And they all agree.

So we've solved the problem and validated the answer is correct.

I've done far more than is required. I could just determine the current through R1 in the first part, and R2 in the second. This would give me the two values for Vx that I could just add up. Knowing this voltage the currents i1, i2, and i3 just fall out.

So to do that:

With -5V switched off we have 10V through 220R giving a current of 45.45mA and Vx of 5.455V

With 10V switched off we have -5V through 275 ohms giving a current of -18.18 mA and Vx of -1.364V

Now adding the two Vx values together we get 5.455 + -1.364 = 4.091V

Thus

i1 = (10 - 4.091)/100 = 59.09 mA
i2 = (-5 - 4.091)/200 = -45.45 mA
i3 = (0 - 4.091)/300 = -13.63 mA

Sorry for the delay in doing this.

#### BobK

Jan 5, 2010
7,682
Very nice explanation Steve, I learned something new.

Bob

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Thanks Steve. I think I understand. If I try to explain it the way I understand it, can you check that I've got it?

You turn off all the external voltage sources by replacing them with shorts to 0V, then you turn one on at a time (separately, not cumulatively) and do a separate calculation for each case.

In each case, all the resistors that are connected to the voltage sources that are disabled are effectively in parallel, and form a single resistance connected to 0V. The resistor(s) that are connected to the enabled voltage source supply current (positive or negative) into the circuit, and form a voltage divider with the paralleled resistors that connect to 0V.

Since there are only two non-zero voltages, there are only two cases. You calculate the currents drawn from the voltage sources, and/or the node voltages, for each case, and add the calculated values of each type together to get the values for when all the voltage sources are enabled.

How did I do?

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Yeah, that seems to describe it pretty well. The disabled voltage sources are simply shorted, not necessarily to gnd though.

Remember that this is a simple case. The voltage sources do not need to be connected to a common ground as these are.

The technique which uses the impedance of the voltage divider is probably the simplest solution, but you can see that all the calculations are suspiciously similar.

The practical benefit of this technique is that it often simplifies the topology of the circuit allowing you to group elements in series and parallel where they were not in this topology originally.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Yeah, that seems to describe it pretty well. The disabled voltage sources are simply shorted, not necessarily to gnd though. Remember that this is a simple case. The voltage sources do not need to be connected to a common ground as these are.
Ah, so this method works where voltage sources are fully isolated from each other. That's an advantage over the Thevenin equivalent method, which assumes all voltage sources are galvanically connected. That's usually the case but not always.
The technique which uses the impedance of the voltage divider is probably the simplest solution, but you can see that all the calculations are suspiciously similar.
Yes I noticed that. It is similar to the Thevenin equivalent method - both use the parallel resistor formula and voltage divider calculations.
The practical benefit of this technique is that it often simplifies the topology of the circuit allowing you to group elements in series and parallel where they were not in this topology originally.
Yes, I can see that. It also works with isolated voltage sources, and it would often involve less calculations than the Thevenin method.

Thanks for taking the time to explain it Steve #### pyrohaz

Oct 28, 2012
33
Thats absolutely fantastic! Thank you very much for all of the help from all of you, i'm so happy that I can do this! I can safely say this problems solved, cheers! #### 24Volts

Mar 21, 2010
164
KrisBlueNZ,

I see how you go the 75ohms by treating R1 and R3 in parallel, but can you please explain how you got the 7.5V???

Thanks

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
I see how you go the 75ohms by treating R1 and R3 in parallel, but can you please explain how you got the 7.5V???
Sure. Pretend R2 (200) isn't there. R1 (100) and R3 (300) form a voltage divider that's connected between +10V (on the R1 side) and 0V (on the R3 side). From there, you can use the voltage divider formula, or you can just notice the ratio of the two resistances, which is 1:3. That means that 1/4 of the total voltage appears across R1, and 3/4 of the total voltage appears across R3. 3/4 of 10V is 7.5V. Does that make sense?

#### Laplace

Apr 4, 2010
1,252
Abandon all these tricky methods and focus on the fundamentals - node equations. What could be more natural than calculating the node voltage? With the node voltage, V, you get the voltage across each resistor and therefore the current through each resistor. When you debug a circuit, what instrument do you use first? A voltmeter! So why would you derive anything other than the node voltage, the same V that would be measured? Plus node equations are the easiest ones to write. You should be able to write node equations in your sleep, but the attached node equation for this circuit should not put you to sleep.

#### 24Volts

Mar 21, 2010
164
Sure. Pretend R2 (200) isn't there. R1 (100) and R3 (300) form a voltage divider that's connected between +10V (on the R1 side) and 0V (on the R3 side). From there, you can use the voltage divider formula, or you can just notice the ratio of the two resistances, which is 1:3. That means that 1/4 of the total voltage appears across R1, and 3/4 of the total voltage appears across R3. 3/4 of 10V is 7.5V. Does that make sense?

Yes thanks for the response, it absolutely makes sense. One other thing though if I may ask, if we pretend R2 isn't there, why are we calculating total resistance of R1 and R3 as though they were in parallel, afterall when we pretend that R2 is not there, aren't R1 and R3 in series and total up to 400 ohms instead of 75 ohms??

24V

#### 24Volts

Mar 21, 2010
164
Abandon all these tricky methods and focus on the fundamentals - node equations. What could be more natural than calculating the node voltage? With the node voltage, V, you get the voltage across each resistor and therefore the current through each resistor. When you debug a circuit, what instrument do you use first? A voltmeter! So why would you derive anything other than the node voltage, the same V that would be measured? Plus node equations are the easiest ones to write. You should be able to write node equations in your sleep, but the attached node equation for this circuit should not put you to sleep.

Thanks Laplace, I will keep this in mind!

One question though:
You are using KCL, so isn't the sum of each branch current equal to the total current... so why does the formula equal to 0? If you make it equal to 0 then shouldn't it be:

V-(-5)/200 - V-10/100 - V/300 = 0

?

24V

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