Here is the circuit I am trying to analyse.


simulate this circuit – Schematic created using CircuitLab

Now I know generally how to do this.

Loop 1 we can call the RIGHT one and Loop 2 we will call the LEFT one.

Loop 2 is clearly 2A because it has a current source.

Loop 1 is 30/10 = 3A

and the voltage at A is clearly 12V because V = 3A*4

But how in the world do I find the Node voltage at B?

  • \$\begingroup\$ Just follow the path from A to B. A = 12 V, across R3 we get ? V because the current is ? A. Then subtract 5V of V1 then add the voltage across R1 (minding the direction of the current and thus the polarity of V(R1) ) then you're at B. \$\endgroup\$ – Bimpelrekkie Nov 25 '15 at 13:48

This is a trick question :'(

It sounds like you are an early student of electrical engineering and, as such, this question is a bit unfair.

The circuit you have drawn is essentially two separate devices and they are connected together by only one wire.

The two devices are:

  • RIGHT: R4, R5, V2
  • LEFT: R1, R2, I1

You see, your reference point (the down-triangle or "Ground") is on the right device (network) and the other device is floating (not referenced to "0 Volts" at any point).

The analysis

The goal, then, is to determine the relationship of LEFT to ground. The key observation is that no current can flow between LEFT and RIGHT at steady state. There is only one wire between LEFT and RIGHT, so there is no return path for the current.


  • You can solve for A using only R4, R5, and V2. It's a simple voltage-divider.
  • The voltage at A is 5V higher than the voltage at the LEFT device's connection to V1 because there is no current in the connecting path, but a 5V drop due to V1.
  • You can solve for the voltage V1(-) to B because you know R1's resistance and R1's current (must be the same everywhere in the loop = I1 = 2A).
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  • \$\begingroup\$ That makes sense! Thank you! and yes this is my first circuits class. \$\endgroup\$ – user125621 Nov 25 '15 at 14:19

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