What are the exact conditions, in terms of graph structure, where KVL can be applied

kirchhoffs-laws

This might be a really dumb question, but …

My understanding of KVL (Kirchoff Voltage Law) is that when you sum
all voltages around a circuit "loop", then the sum is zero.

Fair enough, but in every textbook I've read so far, I have yet to find
a clean, an more importantly formal definition of what a "loop" is.

Is a "loop" any cycle in the circuit graph ?

For example, in the circuit below, can I apply KVL to the left loop, in spite
of it having a short to ground in the middle ?

Is it correct to write v1 + v2 + v3 + v4 = 0 even when v3 is shorted to
ground ?

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Best Answer

There is a misconception here that needs to be cleared up quickly.

When one does KVL 'round a loop one is summing the voltages across the circuit elements in the loop.

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However, you've marked node voltages on your schematic which are, by definition, the voltage from that node to the ground (datum, common, zero reference...) node.

To make it clear, the voltage you've denoted V1 is the sum of the voltages across R1 and R3. This is a consequence of KVL.

So no, you cannot meaningfully sum the node voltages and set them to zero. That would be a misapplication of KVL.

Here are some examples to consider:

By KVL, the voltage across R1 is (V1 - V2).

By KVL, the voltage across R2 is V2 (V3 = 0).

By KVL, the voltage V1 is the sum of the above voltages: V1 = (V1 - V2) + V2

Here's yet another way to look at it.

The voltage V1 is measured by placing the red lead of the voltmeter at node 1 and the black lead at the ground node.

The voltage across R1 is measured by placing the red lead at node 1 and the black lead at node 2.

The voltage across R2 is measured by placing the red lead at node 2 and the black lead at node 3 (the ground node).

By, KVL, the measured voltage across R1 plus the measured voltage across R2 must equal the measured voltage across V1.

Related Solutions

Electronic – Learning loops, Kirchhoff’s Laws

I think you may have an issue with one of the signs. Going around the loop, I get:

12 - 4 i - 2 Vo + 4 - 6 i = 0

Looks like you got the sign on the 6i term flipped.

So, if we take Vo = -6 i, you can calculate:

12 - 4 i + 12 i + 4 - 6 i = 0
16 + 2 i = 0
i = -8
Vo = 48

If you're stuck on understanding how to write this equation, the way to remember it is 'what goes up must come down'. Sum up the voltage across each element as you go around the loop. Since you go into the - side of the 12 volt source, you gain 12 volts. Then you pass through a resistor, and you lose 4 i volts. Then you go the wrong way through the dependent source and you lose 2 Vo volts. Then you go through the 4v source and gain 4 volts. Then you go through the final resistor and lose 6 i volts. You're back where you started, so all of that adds up to zero. Then you figure out what Vo is - if the current is flowing in the direction of the arrow, it will create a voltage drop in opposition to how Vo is marked, so Vo = -6 i. After that, finding i is just algebra.

Electronic – Question about thevenin circuit analysis with voltage across voltage source

This is how i think i would solve the problem:

I don't see it necessary to add a 1V excitation source since there already a 9v in the circuit. The trick when analyzing circuit with dependent sources is to avoid the Short circuiting you would normally do when having a circuit with independent sources alone:

_Things to note that may help

Focus on node A:

define a current from the 9v source direction to node A and also define a current from 6ohm to A

_not Vx =Va

9 - Vx = I1

Vx/6 = I2

and i reckon Vab = 2Vx..

Hope this helps you!

Best Answer

Related Solutions

Electronic – Learning loops, Kirchhoff’s Laws

Electronic – Question about thevenin circuit analysis with voltage across voltage source

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