enter image description here

So supposedly there is a potential difference between those two rods when the circuit is open. Does this mean theres an electric field present here such that ∫Edl≠0 ?

If I put some random loop of wire in the vicinity (I'm not thinking of Faraday's law, instead just a simple line integral acting on charges), will current start flowing "because theres a potential difference"?

Why does KVL apply to open circuits and how exactly does it?

Is it only true for paths close to the circuit?

Can I take trajectory that goes all the way to infinity and back and still apply KVL?

Can I pick a random trajectory that ends where it started and even if theres a gap, assume that the total voltage drop was the one provided by the voltage source?

  • \$\begingroup\$ I have never heard of KVL applying to open loops. Only closed loops. Perhaps if you imagine those two leads of the open loop are a resistor with infinite resistance? Technically, even air has a finite resistance. Imagine that ? in the photo is a 9V battery, then the open voltage across the terminals would be... 9V. \$\endgroup\$ – Justin C Mar 13 '15 at 4:21
  • \$\begingroup\$ Usually if theres an open switch the common jargon is to say " since theres no current in the circuit then all the voltage has gone in between the gap" For some reason people do apply KVL to the open circuit. \$\endgroup\$ – DLV Mar 13 '15 at 4:28
  • 1
    \$\begingroup\$ Unless I'm mistaken, a potential difference will not induce current in an external loop. For that you need a varying magnetic field, which requires alternating current or varying direct current, but your open circuit has no current flow, therefore it generates no magnetic field. \$\endgroup\$ – Justin C Mar 13 '15 at 4:28
  • 1
    \$\begingroup\$ Voltage potential is exactly that, a potential. It would be like seeing a glass of water on a table and saying the water has a potential to spill to the floor. Connecting the terminals with wire would be like knocking the glass over. Justin is right about the magnetic field aspect. Kirchoffs laws apply to any circuit with a loop or junction. An open voltage source has neither. \$\endgroup\$ – I. Wolfe Mar 13 '15 at 4:33
  • \$\begingroup\$ @david. Like I said, imagine a resistor with infinite resistance, and you can create a closed loop. Another way to approach this is the wire connecting either end of said switch has 0 amps and thus 0 voltage drop. If one end of that wire is connected to a potential of 9V, then the other end of that wire (connected to the switch) has still the same potential of 9V. \$\endgroup\$ – Justin C Mar 13 '15 at 4:37

KVL can be more generally stated as: the electric field is a conservative vector field. This can be expressed by the path integral: \begin{align} \int_{p1}^{p2} \vec{E} \cdot d\vec{l} = V_{p2} - V_{p1} \end{align} I.E. the integral is "path independent", and the result is the same no matter what path you take (so long as you start at p1 and end at p2). This can be 0, but it doesn't have to be. If \$p1=p2\$ (taking a closed path), \begin{align} \oint \vec{E} \cdot d\vec{l} = 0 \end{align}

This is why KVL even applies: KVL states that in any closed loop, the voltage across it is 0. It doesn't mean that the voltage must be 0 between any two points on that loop. It doesn't matter if there is any conductive path between two points. Having an ideal wire is just an extra statement that two distinct points are at the same voltage.

| improve this answer | |

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.