Difference between revisions of "Chapter 26 Problem 49"

From 105/106 Lecture Notes by OBM
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==Solution==
 
==Solution==
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===(a)===
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At <math>t=0</math> capacitor is "hungry" for charge, and can be considered as a short. Thus the simpler circuit is simply a circuit with capacitor removed.
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<math> R_{\textrm{eq}}=R+\left( \frac{1}{R} + \frac{1}{R} \right)^{-1}=\frac{3}{2}R </math> <math>\rightarrow I_1=\frac{\mathcal{E}}{R_{\textrm{eq}}}=\frac{2\mathcal{E}}{3R}</math>
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<math>\rightarrow I_2=I_3=\frac{1}{2}I_1=\frac{\mathcal{E}}{3R}</math>
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===(b)===
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At <math>t=\infty</math> capacitor is "full" for charge, and no current will flow through it. Thus the simpler circuit is simply a circuit with the branch with capacitor removed.
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<math> R_{\textrm{eq}}=R+R=2R </math>
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<math>\rightarrow I_1=I_2=\frac{\mathcal{E}}{R_{\textrm{eq}}}=\frac{\mathcal{E}}{2R}</math>
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<math>I_3=0</math>
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===(c)===
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At <math>t=\infty</math> capacitor is "full" for charge, and no current will flow through it. Thus the voltage drop through .

Revision as of 00:12, 31 March 2019

Problem

The RC circuit

In this circuit all resistors have the same resistance R. At with the capacitor C uncharged, the switch is closed.

(a) At the three currents can be determined by analyzing a simpler, but equivalent, circuit. Identify this simpler circuit and use it to find the values of , and at

(b) At the currents can be determined by analyzing a simpler, equivalent circuit. Identify this simpler circuit and implement it in finding the values of , and at

(c) At what is the potential difference across the capacitor?

Solution

(a)

At capacitor is "hungry" for charge, and can be considered as a short. Thus the simpler circuit is simply a circuit with capacitor removed.

(b)

At capacitor is "full" for charge, and no current will flow through it. Thus the simpler circuit is simply a circuit with the branch with capacitor removed.

(c)

At capacitor is "full" for charge, and no current will flow through it. Thus the voltage drop through .