3
$\begingroup$

I always see that Kirchhoff's voltage law either stated as

  1. The algebraic sum of the potential difference in any closed loop is equal to zero

or

  1. The algebraic sum of the component voltage drops in a closed loop is equal to the algebraic sum of the emfs in the loop.

Now from the first way stated, I interpret this to mean, The sum of voltage rises and drops around any closed loop in a circuit is equal to zero. But is it correct to state the law as the algebraic sum of the voltage rises equals to the algebraic sum of the voltage drops in a closed loop?

$\endgroup$
1
  • $\begingroup$ I think your interpretation of the second formulation is correct, so can you explain why you have doubts about your first formulation? $\endgroup$
    – CWPP
    Oct 10, 2021 at 19:32

1 Answer 1

3
$\begingroup$

Now from the first way stated, I interpret this to mean, The sum of voltage rises and drops around any closed loop in a circuit is equal to zero. But is it correct to state the law as the algebraic sum of the voltage rises equals to the algebraic sum of the voltage drops in a closed loop?

Both statements, which I assume address statement 1 of Kirchhoff' voltage law, are correct and essentially equivalent.

Your first statement given mathematically is

$$\sum V=0$$

Stated this way the term "algebraic sum" is essential since voltage rises are assigned positive values and voltage drops are negative values, requiring an algebraic sum.

Your second statement mathematically is

$$\sum V_{rises}=\sum V_{drops}$$

Written this way, it isn't even necessary to say "algebraic sum" in a verbal description of the equation, because all values of the voltage changes on each side of the equation are positive. Then

$$\sum V_{rises}-\sum V_{drops}=0$$ and

$$\sum V_{drops}-\sum V_{rises}=0$$

In either case the important thing is to be consistent in applying the rules for assigning voltage polarities across the circuit elements, which establishes what is a rise and what is a drop.

Regarding statement 2 of the law, I don't much care for it. It could imply that as one moves around a circuit loop all the voltage changes across non emf elements are drops and all emfs are rises. It depends on the direction chosen for the loop current (which is often initially arbitrarily assigned). What's more, a circuit element (e.g. resistor) that is in common between two circuit loops may contribute a voltage rise or drop in one loop, depending on the directions of the currents in the loops.

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.