The Impact of R5 on Req Calculation

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The addition of the R5 resistor in the circuit allows R2 and R3 to be included in the equivalent resistance (Req) calculation because it removes the short circuit created by the capacitor. Initially, with the capacitor acting as a short, R2 and R3 are effectively bypassed, resulting in a simplified current equation through R1. To analyze the circuit, the capacitor should be replaced with a wire to represent its short circuit behavior at time t=0, when it is assumed uncharged and has zero voltage. If the capacitor had an initial charge or if the analysis were at a different time, differential equations would be necessary instead of a simple short circuit replacement. Understanding these principles is crucial for accurate circuit analysis.
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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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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.
 
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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).
 
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