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It’s the first time I come across a problem like this. This is the circuit given. enter image description here I want to find the voltage difference across the capacitor at steady-state before the switch closes.

From what I’ve learn at SS we consider the capacitor as an open circuit. Therefore the bottom line connects the circuit on the left (current source ) with the circuit on the right (voltage source ) .

I’m specifically interested in what happens at the node bellow 5kΩ. Since it connects with the voltage source part of the current coming from the resistance has to go towards the right circuit. But that current never goes back to the current source. I can’t use any of my techniques (node or loop analysis ). It might be possible but this circuit seems a bit odd for me to follow my usual steps.

Well, the voltage drop across the 5kΩ is 50 Volts. If the current goes to the right there is another voltage drop of 4V so the total Voltage difference across the capacitor is 54?

Does the current source provide power to the right current ? What is the potential before entering the current source ?

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  • \$\begingroup\$ There is no current other than in the leftmost and rightmost loops in the steady state with the switch open. So nothing "splits" here \$\endgroup\$ – Eugene Sh. Nov 17 '17 at 21:35
  • \$\begingroup\$ I was thinking the current might go towards the right since there is a path to a lower potential. Why doesn’t that happen ? Well the current re-entering the current source would no longer be at its initial values of course.. I don’t know , something just doesn’t fit. Since there is no exchanging current the potential below the current source is at 6 volts. Correct ? Maybe that’s what’s confusing me. It always went back to zero potential from what I’ve seen until now in my studies. \$\endgroup\$ – John Katsantas Nov 17 '17 at 21:41
  • \$\begingroup\$ Can you update the schematic with names for the nodes so we can discuss it and you'll know what we're talking about? \$\endgroup\$ – The Photon Nov 17 '17 at 21:43
  • \$\begingroup\$ Lower potential? Where? Just draw a ground symbol down there. It's a wire, it's equipotential. \$\endgroup\$ – Eugene Sh. Nov 17 '17 at 21:44
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    \$\begingroup\$ Do you know the cut-set form of KCL? (if not, see here). \$\endgroup\$ – The Photon Nov 17 '17 at 21:47
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There is 54 V across the capacitor at steady state with the switch open:

  • plus side of the capacitor is at +50 V .... 0.01 mA * 5 KOhm

  • negative side of capacitor is at -4 V .... -6 V * 2/3

There is only transient current that flow through the capacitor and the current source and voltage source are completely separate.

When the switch closes then instantaneously there will be -54 V at the junction of the 2k and 4k resistor. Current will flow from the 6 V supply to satisfy the time constant of the RC values as the capacitor discharges to 0 V.

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    \$\begingroup\$ 54 volts. Check polarity of voltage source. \$\endgroup\$ – WhatRoughBeast Nov 17 '17 at 21:54
  • \$\begingroup\$ Oops......you are correct ....fixed. \$\endgroup\$ – Jack Creasey Nov 17 '17 at 22:40
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With only DC sources at steady state you know that Ic=0 so there is no current in the cap or its 0V return path. KVL and KCL rules or simply Ohm's law will then lead to the solution using the R values given and C being open.

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