Difference between revisions of "Chapter 26 Problem 49"
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==Solution== | ==Solution== | ||
+ | ===(a)=== | ||
+ | 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. | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <math>\rightarrow I_2=I_3=\frac{1}{2}I_1=\frac{\mathcal{E}}{3R}</math> | ||
+ | |||
+ | ===(b)=== | ||
+ | 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. | ||
+ | |||
+ | <math> R_{\textrm{eq}}=R+R=2R </math> | ||
+ | |||
+ | <math>\rightarrow I_1=I_2=\frac{\mathcal{E}}{R_{\textrm{eq}}}=\frac{\mathcal{E}}{2R}</math> | ||
+ | |||
+ | <math>I_3=0</math> | ||
+ | |||
+ | ===(c)=== | ||
+ | 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
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 .