First order RC: Find Unknown DC Circuit, given current graph

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SUMMARY

The discussion focuses on analyzing a first-order RC circuit to determine the unknown resistance (R(th)) and voltage (V(th)) given a current graph. The calculations reveal that R(th) is 1 kOhm and V(th) is 6V, derived from the equations involving time constant (Tao) and Kirchhoff's Voltage Law (KVL). The assumption that the capacitor starts uncharged is crucial for the analysis, as it affects the initial conditions of the circuit. The clarification on steady state at t=0 emphasizes the importance of understanding transient responses in RC circuits.

PREREQUISITES
  • Understanding of first-order RC circuit theory
  • Familiarity with Kirchhoff's Voltage Law (KVL)
  • Knowledge of time constant (Tao) in RC circuits
  • Basic capacitor charging equations (Q = CV)
NEXT STEPS
  • Study the transient response of RC circuits in detail
  • Learn about the effects of initial conditions on circuit behavior
  • Explore advanced circuit analysis techniques using Laplace transforms
  • Investigate the impact of different capacitor charge states on circuit performance
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or designing RC circuits will benefit from this discussion.

iharuyuki
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Homework Statement


upload_2015-4-16_14-29-21.png


Homework Equations


[/B]
Tao = R(th) C

The Attempt at a Solution



The unknown DC circuit model consists of a V(th) and R(th)

Tao = (R(th) + 3000) C , total resistance of circuit is R(th) + 3000
0.004 = (R(th) + 3000) (1x10^-6)
R(th) = 1 kOhm

The V(th) should oppose the known source since i(t) goes to 0.
Doing a KVL around the loop clockwise.

-10V + 3000( I ) + I (1000) + V(th) = 0, Voltage across capacitor is 0 since at t=0 it is like steady state.
also, at t=0, I = 0.001 A
then solving gives V(th) = 6V

I get the hunch that the working is wrong but the answer is right by coincidence. Would the working be correct?

Thank you very much.
 
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It's basically correct. The problem statement doesn't explicitly say the capacitor starts out uncharged, but that's a reasonable assumption if nothing else is given.

Your wording "at t=0 it's like steady state" isn't all that clear: steady state is clearly not the case at t=0.
What happens at t=0 follows from a second relevant equation: ##Q = CV## which means that at t=0 the voltage drop over the capacitor is zero (-- provided it's uncharged to begin with!).
 

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