1
$\begingroup$

Imagine a 1.5V battery with both terminals connected securely via a paperclip.

According to KVL, the sum of voltage drops in the circuit (total energy dissipated across circuit elements) must equal the energy afforded by the supply voltage (take for instance the 1.5V AA battery)

How does one interpret KVL in this case? I cannot simply add the voltage drops across the circuit elements, because there are none. I also know that 1.5 volts must get dropped over the circuit, unless KVL doesn't apply.

Closest thing I could find to being the answer was this: Applying Kirchoff voltage law to a short circuit, however it wasn't satisfying.

My guess is that the wire itself must drop all 1.5V and does so through heating, but who knows I could be wrong. Perhaps in this setup the battery is a "short circuit" and KVL doesn't apply?

$\endgroup$
  • $\begingroup$ There are ideal circuits where KVL leads to a contradiction. For example, if an ideal 1.5V voltage source and an ideal wire are connected together, KVL yields $1.5V =0V$ and the current through is undefined ('infinite'). $\endgroup$ – Alfred Centauri Sep 6 '16 at 1:11
2
$\begingroup$

How does one interpret KVL in this case? I cannot simply add the voltage drops across the circuit elements, because there are none.

Well, there is one: The paper clip.

Let's have a try with the Kirchhoff's voltage law: $$\sum V=0 \quad\Leftrightarrow\quad V_{\text{battery}}-V_{\text{paper clip}}=emf-V=0 \quad\Leftrightarrow\quad V=emf$$

Indeed, the voltage drop equal to the battery potential difference ($emf$) must happen in the paper clip. (A tiny amount is though lost internally inside the battery from internal resistance and ought to be included in that equation as well.) If it has a very small resistance, it just acts as a very small resistor. Ohm's law says:

$$V=RI$$

The smaller the resistance, the bigger the current $I$ that flows. This is not so much a problem - the problem is rather that energy dissipation (power $P$) follows current quadratically:

$$P=RI^2$$

Linearly growing current according to Ohm's law causes quadratically growing power, or energy dissipation. Result: A conductor of very small resistance will experience a very high energy dissipation. The material heats up... and melts.

My guess is that the wire itself must drop all 1.5V and does so through heating [...]

Exactly.

Perhaps in this setup the battery is a "short circuit" and KVL doesn't apply?

Short circuit: correct. KVL doesn't apply: wrong.

A "short circuit" is just when close-to-zero resistance wires connect terminals directly and burn. Like the paper clip. With a paper clip of high enough resistance to not melt, it would not be called a short circuit.

$\endgroup$
  • 1
    $\begingroup$ A battery will also have some internal resistance, so it is not true that all voltage is dropped across the wire. $\endgroup$ – fibonatic Sep 5 '16 at 23:59
  • $\begingroup$ @fibonatic Point taken. Edit added $\endgroup$ – Steeven Sep 6 '16 at 0:03
  • $\begingroup$ @efreezy It was my pleasure. $\endgroup$ – Steeven Sep 6 '16 at 0:16
  • $\begingroup$ @Steeven if you could +1 my question, I would earn the much coveted badge "Student"! $\endgroup$ – efreezy Sep 6 '16 at 0:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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