Solving Circuits: Fig. 27-65 Analysis

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KTiaam
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Homework Statement

In the circuit of Fig. 27-65, = 1.2 kV, C = 6.5 μF, R1 = R2 = R3 = 0.73 MΩ. With C completely uncharged, switch S is suddenly closed (at t 0). At t = 0, what are (a) current i1 in resistor 1, (b) current i2 in resistor 2, and (c) current i3 in resistor 3? At t = ∞ (that is, after many time constants), what are (d) i1, (e) i2, and (f) i3? What is the potential difference V2 across resistor 2 at (g) t = 0 and (h) t = ∞? (i) Sketch V2 versus t between these two extreme times.

The Attempt at a Solution


i have loop equations down as:
i1 = i2 + i3
E - i1R -i2R = 0
i2R - i3Ri don't understand how they get the answer
i1 = 2/3(v/r)

Ive also attached the steps shown from the solution manual:
thanks for all you help
 
Attachments
  • Circuit schematic of Fig. 27-65: a battery labelled 𝓔 in series with resistor R1, a switch S, then resistor R2 and resistor R3 with capacitor C in the branch to the right.
    Screen Shot 2014-04-20 at 11.57.12 AM.png
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  • Handwritten-style typeset solution: i1 = 2E/(3R) = 2(1.2x10^3 V)/(3(0.73x10^6 Ohm)) = 1.1x10^-3 A; i2 = E/(3R) = 1.2x10^3 V/(3(0.73x10^6 Ohm)) = 5.5x10^-4 A, and
    Screen Shot 2014-04-20 at 12.06.15 PM.png
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  • Like
Likes   Reactions: ilker29
I don't know why they went so far as to solve simultaneous equations for this problem. Really, since the at the times of interest the capacitor either looks like a short circuit or an open circuit, the result is a simple voltage dividers to find the potential at the switch. With that potential in hand the currents in the resistors is straightforward Ohm's law.

If you take the voltage divider approach you should find their expression for i1 quite easily.