2
$\begingroup$

There are many books which relate KVL with law of conservation of energy and I'm having trouble understanding how they are related so can anyone explain their relation to me, concept wise and if possible arthimetically?

$\endgroup$
2
  • 1
    $\begingroup$ My answer here might help physics.stackexchange.com/q/172284 (Particularly the last paragraph). If not I will write an answer. $\endgroup$
    – ChrisM
    Mar 25, 2015 at 18:09
  • $\begingroup$ @chris2807, Yes, but the problem is why does it have to be around a loop $\endgroup$
    – Socre
    Mar 25, 2015 at 18:28

2 Answers 2

2
$\begingroup$

Kirchoff's voltage law.

Quoting Wikipedia,

The directed sum of the electrical potential differences (voltage) around any closed network is zero, or:

More simply, the sum of the emfs in any closed loop is equivalent to the sum of the potential drops in that loop, or:

The algebraic sum of the products of the resistances of the conductors and the currents in them in a closed loop is equal to the total emf available in that loop.

Why it has to be around a loop?

When you travel a complete loop, you are back to the initial point from where you started. Since, the potential of the point where you started will remain constant, hence you can infer that:

The change in potential around any closed loop is zero. Otherwise, the initial and the final points (which are same) would have different potentials if change in potential is not equal to zero.

Otherwise, if there is a potential gain or loss, we could travel round and round the loop, gaining / losing potential which would sum up to an infinite series and the law of conservation of energy would be violated.

$\endgroup$
0
0
$\begingroup$

Consider the case where we just have a battery and a resistor in series and nothing else. Lets say the battery gives the electron some amount of energy $U$. Now look at one electron as it travels around the circuit. Let's say the battery gives the electron some amount of energy $U$.

If the electron still had some energy left when it arrived back at the battery $\alpha U$ where $0<\alpha<1$ it would receive another amount of energy $U$ from the battery so now its energy would be $U+\alpha U$.

So you can see we will run into a problem here as the electron will keep retaining a fraction of energy $\alpha$ each trip round the circuit. We have effectively set up a geometric series which we could sum to infinity to find the steady state energy of the electron and we would find that the energy is greater than the initial energy the battery had to give (do this for yourself)!

If you consider the opposite case it is obviously nonsensical, the electron can not lose more energy on each trip of the circuit than the battery had to give initially.

So the only thing that would make sense physically is the electron losing exactly the amount of energy the battery had to give in the first place. This leads directly to KVL.

This obviously isn't a proof but hopefully gives you a good intuitive argument as to why it must be this way.

$\endgroup$

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