Engineering Find di/dt and dr/dt for a DC circuit with indepedent current source

AI Thread Summary
The discussion focuses on solving a DC circuit problem involving an independent current source, specifically finding diL(0+)/dt, dvC(0+)/dt, and dvR(0+)/dt. The initial conditions indicate that before t=0, the circuit behaves as an open circuit due to the absence of the 4A current source, with the inductor acting as a short circuit and the capacitor as an open circuit. At t=0+, the inductor current remains at -6A, leading to the conclusion that diL/dt is 0 since the current does not change abruptly. For the resistor, it is established that there is no current flow at t=0+, resulting in dvR/dt also being 0. The discussion highlights the importance of understanding circuit behavior at the transition point and the implications of ideal components.
Xhendos
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Homework Statement
For the circuit in Fig. 8.7, find: (a) i L (0 + ), v C (0 + ), v R (0 + ),
(b) di L (0 + )∕dt, dv C (0 + )∕dt, dv R (0 + )∕dt, (c) i L (∞), v C (∞), v R (∞).
Relevant Equations
I = C dv/dt
V = L di/dt
Dear PF,

I am trying to solve practice problem 8.2. I am stuggling with part B of the question which asks us to find diL(0+)/dt, dvC(0+)/dt and dvR(0+)/dt.
2021-01-23-123306_909x406_scrot.png
My attempt can be seen below. Before t=0 the 4A current source is not present and thus it is an open circuit. A capacitor acts as an open circuit to DC and an inductor acts as a short circuit to steady-state DC conditions and thus that is why schematic for part A of the question looks like that. Using this schematic I find that iL = -6A and both vC and vR are 0V. This matches the answer of the book.

However, for part B of the question, I did manage to find dv/dt by using KCL at node A and B and using the fact that the inductor cannot change its current abruptly, so for t = 0+ the inductor current is still -6A.
WhatsApp Image 2021-01-23 at 12.19.37.jpeg


However, I am not quite sure how I can find di/dt and dr/dt. Can anyone give me a little hint here? I am stuck on this for quite a while.
 

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Xhendos said:
the inductor cannot change its current abruptly, so for t = 0+ the inductor current is still -6A.
so what does that say about the (initial) ##\displaystyle {di_L\over dt} ## ?
 
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BvU said:
so what does that say about the (initial) ##\displaystyle {di_L\over dt} ## ?
Right! Since the current iL does not change between interval 0- and 0+ the derivative at that point in time is 0! You are right. How about the resistor? Can we proof that current iR is 0 such that there is no potential voltage across the resistor its derivative and no current flows at t=0+ through the resistor?
 
Xhendos said:
there is no potential voltage across the resistor its derivative and no current flows at t=0+ through the resistor?
You found ##V_R(0^+)## in part (a). For the current the capacitor can swallow at ##t=0^+## there is (only theoretically) no limit: the current source is supposedly ideal and pumps in 4A, thereby determining ##\displaystyle {{dV_C\over dt} \left (0^+\right )}## according to what you said yourself:
Xhendos said:
1611442195524.png
The righthand side of the circuit doesn't interfere with this (the inductor does not like sudden changes in current).

But now comes the hard bit: ## \displaystyle {{dV_R\over dt}\left (0^+\right )}##.
Charles' book answer is 0. What do you think :smile: ?

Xhendos said:
1611443745557.png
This assumption is indeed necessary.
But, different from 8.1, I don't see it in the problem statement :rolleyes:...
 

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