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Circuit

In this circuit. What is the current passing through R1?

I was thinking about looking at the left and right sections separately and then just add both currents to get the total current. But then I realized that for eg the 50 ohm resistor and R1 are not parallel, but they are not in a series either.

How should I approach a circuit problem like this?

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    \$\begingroup\$ Do you have a ground in this circuit? \$\endgroup\$
    – Syed
    Mar 26, 2022 at 15:09
  • \$\begingroup\$ I guess not, the picture is a screenshot of the circuit. \$\endgroup\$
    – Gripen
    Mar 26, 2022 at 15:13
  • \$\begingroup\$ @DKNguyen, OP's diagram might remind you of a transmission line problem, but what it actually is is whatever OP thinks it is. We don't know whether OP has ever heard of "transients" or "transmission lines." \$\endgroup\$ Mar 26, 2022 at 15:30
  • \$\begingroup\$ @SolomonSlow I doubt it actually is a transmission line problem since it is incomplete for that. But in a real circuit if you actually wanted to figure out R1's current then it would be. It's much more likely to be a regular problem to try and trip up beginners. \$\endgroup\$
    – DKNguyen
    Mar 26, 2022 at 15:33

2 Answers 2

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This is a transmission line problem if you really want to figure out the momentary/transient charge that flows in R1 when you connect up the power supplies to the circuit and would involve adding L and C to model those real world effects.

But as drawn, the circuit is ideal and in a vacuum so no current flows through R1 because there is no closed circuit for it to flow. Since no current flows through R1 then the voltage drop across R1 is 0V via V=IR. That means the potential on both sides of R1 are the same. That might take some time to wrap your head around if you're new to this.

So if you took a volt meter and measured the voltage between the two negative terminals of each source you would actually measure 40V (or -40V depending on which way you stuck the red and black probe).

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  • \$\begingroup\$ Thanks for clear explanation! \$\endgroup\$
    – Gripen
    Mar 26, 2022 at 15:30
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    \$\begingroup\$ @Gripen, DNK said that the potential on either side of R1 is the same, but note! That potential also is undefined. Voltage only is ever truly defined between two different points in the circuit. If somebody says "Voltage at" some point, the conventional meaning is the voltage between that point and the node called "ground." But there is no ground shown in your circuit, so... \$\endgroup\$ Mar 26, 2022 at 15:33
  • \$\begingroup\$ @SolomonSlow Yes, there's potential and potential difference (voltage). So saying the potential at two points is the same as saying that the potential difference is zero but you can't actually ascribe a number to just the potential at one of the points. You can only state a number for the potential difference between two points. By sameness, it also means that if you chose a point in the circuit to be defined as 0V which you measure the voltage at any other single point with respect to as a reference, then whatever you measured between that point and each side of R1 would also be the same. \$\endgroup\$
    – DKNguyen
    Mar 26, 2022 at 15:40
  • \$\begingroup\$ and still have a difference of zero. \$\endgroup\$
    – DKNguyen
    Mar 26, 2022 at 15:40
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Currents flow in loops. No loop - no current. Same loop cannot pass multiple times through the same 2-port passive device. Here, you cannot draw a loop over the circuit elements that includes a voltage source and R1. So the current has to be zero.

In fact, you can only draw two loops: one through the left battery and the 50 Ohm resistor, another through the right battery and 10 Ohm resistor. That's it. The 40 Ohm resistor is not covered by any possible loop, so it's exactly as-if it wasn't there, in terms of the circuit theory at least.

In practice, this resistor would have some effect in more sensitive circuits, and definitely would make a difference in many RF circuits.

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