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I am trying to follow this example problem: enter image description here

I understand the current calculation. What I dont understand is why the 5 kOhm resistors voltage drop does not need to be taken into account when finding v2. The book says v2 is the same as the voltage across the 4 kOhm resistor. Why do you not need to take into account the voltage drop across the 5 kOhm resistor?

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    \$\begingroup\$ Hint: The problem says "under DC conditions". If the circuit is in steady-state what's the current flowing in the capacitor? \$\endgroup\$ – John D Jun 17 at 14:16
  • \$\begingroup\$ @JohnD the current would be 0 since the capacitor is an open circuit \$\endgroup\$ – Stone Preston Jun 17 at 14:17
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    \$\begingroup\$ And what is the voltage drop across a \$5\mathrm{k\Omega}\$ resistor with zero current? Or any finite-valued resistor, for that matter? \$\endgroup\$ – TimWescott Jun 17 at 14:49
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Under DC conditions, the capacitor will be as fully charged as it will ever be.

That implies that the current through the capacitor has dropped to zero, so the current through the 5K resistor is also zero. So the voltage across the resistor is zero.

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  • \$\begingroup\$ alright, so why wouldn't the voltage across the capacitor be 0 then, since as you said the current through the capacitor is 0 as well. That is the part that is confusing me. \$\endgroup\$ – Stone Preston Jun 17 at 14:22
  • \$\begingroup\$ Remember that DC cannot pass through the capacitor, so once v2 is fully charged it's like the 5k resistor isn't connected to anything on one side. You can see that in the second diagram. But v2 had to be fully charged for that state to occur, so the voltage difference is still present. It's like a battery, even if no current is flowing, there is still a voltage difference across it, no? \$\endgroup\$ – hekete Jun 17 at 14:30
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    \$\begingroup\$ @StonePreston because it's not a resistor, so Ohm's law doesn't apply. If you apply a voltage across a capacitor, it charges until it reaches the same voltage as the one you are applying. Then the current stops. \$\endgroup\$ – Simon B Jun 17 at 14:39
  • \$\begingroup\$ alright perfect. Now I understand. Thank you guys \$\endgroup\$ – Stone Preston Jun 17 at 14:43

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