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image1

I want to find v1. So I use KVL down the path from v1 to -3V.

I assumed the answer was v1+0.7+-3=0, so v1=2.3

But the actual solution was v1–0.7– - 3=0, so v1=-2.3

Why is this? Is it because the current is going into the voltage source opposite to sign convention? If so, does that mean the -3 terminal at the end should also be treated like a voltage source opposite the sign convention for the purposes of solving? e.g.

image2

So v1 + (-) (+0.7 V) + (-) (-3 V)=0, so v1–0.7+3=0 giving v1=-2.3 ?

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    \$\begingroup\$ V1 is 0.7 volts higher than -3 volts i.e. V1 = -2.3 volts. Don't use KVL, use your brain. \$\endgroup\$
    – Andy aka
    May 2, 2020 at 16:40
  • \$\begingroup\$ But if I was given a question that explicitly asked me to use KVL, would my understanding be correct? \$\endgroup\$
    – johnwick
    May 2, 2020 at 16:50
  • \$\begingroup\$ You have expressed several different versions of your understanding so yes and no (version dependent). \$\endgroup\$
    – Andy aka
    May 2, 2020 at 17:07

2 Answers 2

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To apply KVL, add the load between v1 and ground to your schematic. Then make a directed sum of the potential differences; if I follow your direction of I, I get:

+0.7 - 3 - v1 = 0

That's because the direction over the load would go upwards to get back from your GND to the point labeled v1 in your schematic.

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  • \$\begingroup\$ So if the question did not provide a load or ground I would just add a hypothetical one in for the purposes of using KVL? \$\endgroup\$
    – johnwick
    May 2, 2020 at 17:25
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    \$\begingroup\$ Voltages don’t mean anything unless they are referenced to something. \$\endgroup\$ May 2, 2020 at 18:54
  • \$\begingroup\$ @johnwick The rail voltages shown in the diagram are already assumed to be "with reference to ground." You could, if it helps you, add two voltage sources -- each with one end tied to ground and the other end tied as appropriate to the "loose ends" you see labeled with a voltage. People, quite often, actually remove the voltage sources themselves and just label the "ends" to make it "simpler." But "simpler" is in the eye of the beholder, I suppose. \$\endgroup\$
    – jonk
    May 2, 2020 at 18:56
  • \$\begingroup\$ @johnwick Applying a load to V1 and ground is an incorrect approach. In this example it just works because V1 is connected to ideal voltage sources. If the 0.7V voltage spurce were a diode or a resistor, and the load a 10Ω, then this approach would definitely yield a wrong answer. \$\endgroup\$
    – Huisman
    May 2, 2020 at 20:10
  • \$\begingroup\$ spurce --> source \$\endgroup\$
    – Huisman
    May 2, 2020 at 21:35
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For applying KVL you need a closed loop.
As commented above, it is an incorrect approach to add "random" components to V1 in order to close the loop.
Moreover, it is not needed, because the schematic is already a closed loop.
Like @jonk comments, you should draw the ground symbols of the +3V and -3V sources. Then, it is not hard to find the closed loop (see red dashed box) and apply KVL.

Next, you need to find V1 … but with respect to what?
If it is with respect to ground, KVL is a quite dumb approach as it yields a result you could think of without using KVL.
If you need to find the voltage across the resistor of 10kΩ then KVL makes sense.

schematic

simulate this circuit – Schematic created using CircuitLab

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