I'm studying Circuit Analysis and a sample problem is as follows:

enter image description here

I'm not really sure how to approach the problem. My intuition leads me to believe that the two voltage sources are acting against one another versus stacking their effect.

When I search this problem online, I come across examples of superposition, which breaks the circuit apart, analyzes the individual parts, then brings the parts together in an additive or subtractive manner. Thus far in my studies I haven't learned superposition, thus I feel that I shouldn't need it to find the required values.

I'm not looking for a walkthrough for this problem, just somewhere to start. I've thus far learned KVL, KCL, and Ohm's Law. Apparently KVL should be sufficient for solving this problem, I'm just not seeing it.

What method would I use to find the current? I believe that if I know the current I can solve the remainder of the problem.


4 Answers 4


In this circuit KVL would be enough to solve. Applying KVL would sum all voltages and equal them to zero. Initially taking the given current direction as correct, we have: 5-i-9i-10-10i = 0, then i = -0.25, which tells us the current is in opposite direction regarding the one given in the image.

Found the current, Vab is the voltage across the 1ohm resitor. As current flows from B to A(we know that because we discovered its value in the topic a) ), the voltage Vab is negative, so Vab = -1*0.25 = -0.25V.

Finally,the voltage in node B is GREATER than the voltage in node A. We know that because if you measure the voltage from B to GND you will find Vb = 5 + 0.25 = 5.25V, while Va is 5V because is the source voltage.


Apparently KVL should be sufficient for solving this problem, I'm just not seeing it.

It's as simple as adding up all the voltage drops around the loop, like you'd do in any KVL problem:

$$5\ {\rm V} - i (1\ {\rm \Omega}) - i (9\ {\rm \Omega}) - 10\ {\rm V} - i (10\ {\rm \Omega}) = 0$$

This is one equation with one unknown, and it's linear, so it will have a single unique solution.


All the components are in series so you can re-arrange in any manner providing you keep the voltage polarities the same. This means you can have the 10 ohm resistor that is in the ground limb moved up in the top limb. This should make it a lot easier to see the wood from the trees.


This circuit is a single loop with all components in series, so all the components of the same type can be merged into their series combinations. The exception is RA1, since we're interested in the voltage across it - we can't merge it for that purpose. But we can merge it to determine the current in the loop.

  1. Merge RA2+RA3 into RB23.
  2. Merge VB1+VB2 into VC12.
  3. Merge RC23+RC1 into RD123.


simulate this circuit – Schematic created using CircuitLab

  • The current \$i = 5{\,\rm V}/20{\,\Omega} = 1/4{\,\rm A}\$.
  • The voltage across RC1 is \$v_{RC1} = 1{\,\Omega}\cdot 1/4{\,\rm A}=1/4{\,\rm V}\$.
  • Due to polarity of VC12, the voltage at B is higher than the voltage at A.

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