The Impact of R5 on Req Calculation

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Homework Statement
Consider the circuit to the right, with R5 = 119 Ω in series with the capacitor. Once again, the switch has been open for a long time when at time t = 0, the switch is closed. What is I1(0), the magnitude of the current through the resistor R1 just after the switch is closed?
Relevant Equations
I=V/R , Req=R235+R14
So what I know is that without the edition of the R5 resistor R2 and R3 are 0 because the capacitor is short circuiting them. Why does the addition of the R5 resistor cause R2 and R3 to be calculated into the Req for the current equation?
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kuruman said:
Adding R5 makes a difference because R2 and R3 are no longer shorted. Redraw the circuit with the capacitor as a "short" to find the initial Req and hence the initial current through R1 the usual way.

Thank you I kind of understand that, but how do I redraw the circuit with the capacitor as a "short". I don't really understand what that looks like
 
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mabdeljaber said:
Thank you I kind of understand that, but how do I redraw the circuit with the capacitor as a "short". I don't really understand what that looks like
Replace the capacitor with a wire.
 
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gneill said:
Replace the capacitor with a wire.

Thank you! Understood
 
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Maybe we have to explain why the capacitor at time t=0 behaves as a short. The reason is that the capacitor is assumed to be uncharged so its voltage is zero at this time instant. So if you apply KVL for the time instant t=0 you ll see that the resulting equation is like that the capacitor is a short.

If the capacitor had some initial charge or if we were asked for the current at another time instant then you shouldn't replace the capacitor with a short. You will had to solve differential equations then (unless the charge of the capacitor at that time instant was given).