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I got this question in an exam, where I was given an unbalanced Wheatstone bridge and was asked to find the voltage across the capacitor. I can't understand how to apply Kirchhoff's laws for a capacitor, as I'm only given capacitance. I simulated the circuit in Falstad and got the voltage across the capacitor as 2 V. How do I apply KVL to this circuit? It is neither series nor parallel and I don't know how to solve it.

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  • \$\begingroup\$ Since the switch is open, the voltage across any component is zero and no current flows... You might want to explain the purpose of that switch for the question to make sense. \$\endgroup\$
    – Lundin
    Sep 1, 2023 at 13:18

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R2 and R4 make a voltage divider. When there is no load on this voltage divider (i.e. when the capacitor is in steady state), its output (node b) will be (30/50)*45 = 27V.

R1 and R3 also make a voltage divider. Where there is no load on this voltage divider, its output (node a) will be (50/90)*45 = 25V.

The difference between these two is 27-25=2V.

Edit:

I can see how we use the voltage divider rule to soLve it, is there any other method that we can follow to solve the problem, like KVL?

If you ignore the capacitor, there are two independent loops in your circuit (three total). There is a purely resistive loop, and there are two loops that include the power supply.

What KVL tells you is that the voltage around any loop is 0. Or, put another way, that if you go from one node to another in two different ways, the voltage from one node to the other will be the same regardless of the route.

I implicitly used KVL by assuming that the voltage across each voltage divider (across R1 and R3, and across R2 and R4), and across the battery are all equal.

That is as far as KVL, by itself, will get you. You then need to apply Ohm's law to find the currents, through each resistive leg. Then having found the currents through each resistive leg, you need to use Ohm's law again to find the voltages.

One can write that all out, if for example, one is required to do so for an academic assignment or a test. However, voltage dividers are so ubiquitous in electronics, that it should become second nature when one sees two resistances in series, to just apply the voltage divider rule.

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  • \$\begingroup\$ Can you please explain if the R2 and R1 Node parallel? for a neither series nor parallel circuit how does the voltage and current get divided between each branch? \$\endgroup\$
    – Mayur
    Aug 31, 2023 at 16:10
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    \$\begingroup\$ R1 and R2 are not parallel. R2 and R4 are in series if you ignore the capacitor. R1 and R3 are in series if you ignore the capacitor. \$\endgroup\$ Aug 31, 2023 at 17:09
  • \$\begingroup\$ thanks for the response! I can see how we use the voltage divider rule to soLve it, is there any other method that we can follow to solve the problem, like KVL? \$\endgroup\$
    – Mayur
    Sep 1, 2023 at 10:59
  • \$\begingroup\$ I have made an addition to my answer regarding KVL. However, on reviewing the question, I am wondering if what was wanted was the voltage as a function of time. I just gave the final state after the switch is closed. \$\endgroup\$ Sep 1, 2023 at 11:30
  • \$\begingroup\$ Thanks for answering! That cleared my doubts 😃 \$\endgroup\$
    – Mayur
    Sep 2, 2023 at 18:25

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