# Homework Question basic circuit

#### lentim82

Joined Aug 15, 2016
4
please see attached pic. The solution to this was rather lengthy using KVL etc.

Is there a simpler way to get V2?

thanks

#### WBahn

Joined Mar 31, 2012
25,064
We are not mind readers -- since you don't show any work for how you got it using KVL, how can we possibly tell you whether there is a simpler way?

#### DGElder

Joined Apr 3, 2016
351

#### lentim82

Joined Aug 15, 2016
4
sorry, here is the solution provided (see attached). Just wondering if there's a faster/easier way.

thanks

#### DGElder

Joined Apr 3, 2016
351
sorry, here is the solution provided (see attached). Just wondering if there's a faster/easier way.

thanks
That's not KVL, it is KCL.

Try superposition. Solve for current from 1A source first, in each 4 ohm resistor: R2 and R1. Then due to the symmetry of the load and because the left hand source is twice the right hand source, you know that the total current in R2 will be the current in R2 + 2 times the current in R1. Then it is just I*R2 to get V2.

Last edited:

#### WBahn

Joined Mar 31, 2012
25,064
sorry, here is the solution provided (see attached). Just wondering if there's a faster/easier way.

thanks
The solution shown uses KCL (not KVL) and shows the steps involved in getting two equations in two unknowns in gory detail, which is what makes it appear long. With a bit of practice and using a formalized way of applying KCL known as Nodal Analysis, you can write the two equations by inspection as:

V1[1/R1 + 1/R3] - V2[1/R3] = 2 A
-V1[1/R3] + V2[1/R2 + 1/R3] = 1 A

At that point you can easily solve for V2.

There are a few other methods.

Using Mesh Analysis (which is just a formalized way of applying KVL) you can write, by inspection, a single equation for the current flowing in R3 and then use that to find the voltage across R2.

You can also use superposition pretty effectively on this one.

Another option is to use source transformation to reduce the left hand side (the 2 A source plus R1 and R3) into a 4/3 A current source in parallel with a 6 Ω resistor and then combine the two current sources into a single 7/3 A current source. Now use current division to find the current through R2 and then Ohm's Law to find the voltage across it.

V2 = (7/3 A)·[(6 Ω)/(6 Ω + 4 Ω)]·(4 Ω) = 5.6 V

#### lentim82

Joined Aug 15, 2016
4
thank you so much. Can you please exlplain the source transformation method a bit further..

I don't understand how the 2A source becomes 4/3 A in parallel with a 6ohm resistor.

#### WBahn

Joined Mar 31, 2012
25,064
thank you so much. Can you please exlplain the source transformation method a bit further..

I don't understand how the 2A source becomes 4/3 A in parallel with a 6ohm resistor.

Sure.

http://bfy.tw/7EEq

#### lentim82

Joined Aug 15, 2016
4
thank you WBahn I understand now how you get the simplified source transform circuit. (attached sketch)

The last question I have is.. you are able to simply add up those current sources? If they were voltage sources, would you still be able to add them?

Thanks!

#### WBahn

Joined Mar 31, 2012
25,064
thank you WBahn I understand now how you get the simplified source transform circuit. (attached sketch)

The last question I have is.. you are able to simply add up those current sources? If they were voltage sources, would you still be able to add them?

Thanks!
Yes, you can add them because each one is injecting current into the top node. If they were voltage sources you would most definitely NOT add them. In fact, unless they have the exact same voltage output, two voltage sources in parallel result in an undefined voltage on that node because the two voltage sources will fight each other. In the real world voltage sources all have some amount of series resistance which will determine the resulting voltage. But even in the real world, if the two sources differ by too much voltage and the series resistances are too small, the result will be such a large current flow that something bad will happen, ranging from overheated and damaged components to a catastrophic and potentially dangerous explosion.

For current sources, a similar situation exists if they are put in series. They will fight and either the non-ideal limitations of one (or both) will determine the resulting current or, if the two sources are too close to being ideal, there will be a damaging, potentially catastrophic, failure of one or both sources.

#### DGElder

Joined Apr 3, 2016
351 You asked for simpler method, not different method. Using the principle of superposition is simpler.

At first glance you can see a circuit that cries out to be solved by superposition. The symmetric load seen from the right hand source looks like the mirror image of the load as seen from the left hand source. And you have a source that is conveniently 1A and the other which is a simple multiple of 1A. With a little experience you can solve it in your head.

Solve for the right hand source first. By inspection of this current divider, the current in R2 and in the R1+2 ohm combination must be 6/10 A and 4/10 A respectively.

Then solve for the left hand source. The current division in the load is just a reflection of what you saw for the right hand source - but multiplied by 2. So the current in R2 from the left hand source is 2*4/10 A = 8/10 A. Now add the results from both sources. 6/10 + 8/10 = 14/10 = 1.4A.

So the voltage V2 = 1.4A * 4ohms = 5.6V.

Last edited: