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So let's take this circuit:

enter image description here

To find voltage across \$ V_1 \$, you do KVL in loop 1 to find \$ V_1 \$

you say:

\$ 5V + V_1 + V_3 = 0 \$

correct? But why is it assumed that current flows through the 5V supply? Aren't the positive and negative terminals of the battery (5V supply in this case) separated? Current direction defined as flowing from the positive terminal to negative terminal, but let's ignore that, and talk about what happens in reality. Physically, electrons will go from the negative terminal of the 5V supply around the loop to the positive terminal of 5V supply.

So, the 5V supply is really an open circuit, isn't it? Whenever you do KVL or these loop equations, why is it assumed that current flows seamlessly between voltage supplies?

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    \$\begingroup\$ NO a supply is not an open circuit. You might find this cross-post interesting. electronics.stackexchange.com/a/338888/139766 \$\endgroup\$ – Trevor_G Nov 16 '17 at 3:11
  • \$\begingroup\$ your V1 and V3 polarities are incorrect \$\endgroup\$ – jsotola Nov 16 '17 at 3:30
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    \$\begingroup\$ @jsotola The choice of reference directions is arbitrary, and the equation given is consistent with the chosen directions. \$\endgroup\$ – The Photon Nov 16 '17 at 3:50
  • \$\begingroup\$ i disagree. the diagram implies that the voltage at the junction of R1 and R3 is equal to 5V + V1 .... if the polarity of any component of the circuit is not evident at first glance, then such markings should not be introduced into the circuit at the begining \$\endgroup\$ – jsotola Nov 16 '17 at 5:04
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First, Kirchoff's Voltage Law (KVL) applies whether there's current flowing through the branches you traverse or not. In your circuit you could replace R3 with an open circuit, and replace R2 with an arbitrary network or resistors, transistors, etc., and your KVL equation would still hold: \$5\ {\rm V}+V_1+V_3=0\$. It wouldn't be much use for calculating the branch currents because you wouldn't be able to write \$V_3\$ in terms of branch currents, but it would still be a valid equation.

Second, current does flow through the battery. It doesn't flow as electron current in a metal, but as ions flowing through the battery's dielectric.

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  • \$\begingroup\$ >"Second, current does flow through the battery. It doesn't flow as electron current in a metal, but as ions flowing through the battery's dielectric." I'd like you to elaborate on that. How much are we talking about here? Is it like 1/10 of what went from negative terminal originally or ...? \$\endgroup\$ – Jack Nov 16 '17 at 5:46
  • \$\begingroup\$ @Jack Battery chemistry uses terminology all its own. The details can be complicated. I just saw a new report saying some scientists have demonstrated that quantum tunneling also takes place in batteries and that not all of the charge transfers take place at the anode and cathode chemistry boundary layers. That is just how complex all this can really be in practice. For you and me, it's enough to accept that the total effect of all causes is that current flows through a battery. For details how currents actually happen, I recommend "Matter & Interactions," 3rd edition. \$\endgroup\$ – jonk Nov 16 '17 at 6:40
  • \$\begingroup\$ It's also sufficient just to accept that the current in one terminal of a battery equals the current out the other. \$\endgroup\$ – David Schwartz Nov 16 '17 at 8:04
  • \$\begingroup\$ @jonk, wait a sec, NO. Current cannot flow through battery! The battery will explode, wouldn't it? If electrons have access to terminal with no resistance... It's like shorting + and - terminals of battery, no? Plus and minus terminals are isolated! There's physical barrier between them! \$\endgroup\$ – Jack Nov 22 '17 at 3:18
  • \$\begingroup\$ @jack, the current that flows through the battery is not electrons flowing in metal. It's ions flowing in the battery's dielectric. \$\endgroup\$ – The Photon Nov 22 '17 at 3:47

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