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The question that I'm trying to solve is as follows:

In the circuit shown below, the switch has been closed for a long time. a) What is v(0) or the voltage across the capacitor immediately after the switch is opened? b) What is i(0) and i1(0)?

enter image description here

First, I calculated for the voltage before t = 0 using voltage division. I arrived at v(0-) = 4V. Since the capacitor voltage cannot change instantaneously, I assumed that 4V is the voltage at t=0.

Here's the tricky part. I'm asked to find the current at t = 0. By looking at the diagram, I am certain that the 1-ohm resistor is short-circuited, so I simply dismissed that and used Ohm's Law to find i1(0-) using v = 20 V and R = 4 ohms. In the property of short circuit, the i is equal to 0.

Are my solutions for i1 and i legal to begin with? I'm only confident with 4V as the voltage at t = 0.

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    \$\begingroup\$ Is the question asking for currents at \$0^-\$ or \$0^+\$ (i.e. before or after switch opening) ? Also, what do you mean by 1 ohm resistor is short circuited ? after \$t=0\$, it is effectively removed from the circuit by the open switch. \$\endgroup\$ – AJN Jun 18 at 16:50
  • \$\begingroup\$ You need to take into account the voltage across the capacitor at t=0+ (you already found this) when calculating the instantaneous current flow in the 4 ohm resistor. \$\endgroup\$ – Phil Freedenberg Jun 18 at 17:01
  • \$\begingroup\$ @PhilFreedenberg, the diagram has two currents in question--i1 and i. Are they assumed to be equal? \$\endgroup\$ – romeoPH Jun 18 at 17:49
  • \$\begingroup\$ @romeoPH Immediately after the switch is opened, what can you say about i1 and 1? (This is not an assumption; use KCL) \$\endgroup\$ – Phil Freedenberg Jun 18 at 23:39
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Homework guidance:

Nobody shorts the 1 Ohm resistor, its end is opened so its current is stopped at t=0. Before t=0 there were 4A as your voltage division shows.

At t=0 the circuit changes. You have 4V in the capacitor, 20V in the source and a resistor between. The capacitor charges towards 20V starting from 4V. Find the formulas.

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  • \$\begingroup\$ "Before t=0 there were 4A as your voltage division shows." Do you mean 4 V? \$\endgroup\$ – romeoPH Jun 18 at 17:34
  • \$\begingroup\$ 4V means also 4A in the 1Ohm resistor when the resistor is still connected. \$\endgroup\$ – user287001 Jun 18 at 17:35
  • \$\begingroup\$ Is it safe to assume that i1 is equal to i in this case since the 4-ohm resistor and the capacitor are in series (and therefore, have the same current)? \$\endgroup\$ – romeoPH Jun 18 at 17:40
  • \$\begingroup\$ After the switch is opened i1=1, there's no other way left for the current. Being in series means the current is the same literally. \$\endgroup\$ – user287001 Jun 18 at 17:42
  • \$\begingroup\$ What do you mean by i1=1? That the current is the same? \$\endgroup\$ – romeoPH Jun 18 at 17:53

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