RC Circuit + Dependent Voltage Source

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  • #1
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Homework Statement



http://img167.imageshack.us/img167/7747/picture2mg7.png [Broken]

Homework Equations



iL(t) = current through inductor

Using The Step by Step Method:

Assume the solution iL(t) = K1 + K2*e^(-t/tau)
where K1 = iL(infinity) and K2 = iL(0+) - iL(infinity)

so, Assume the solution has the form:

iL(t) = iL(infinity) + [iL(0+) - iL(infinity]*e^(-t/tau)

Now, what I did was I calculated iL(0-) first... I replaced the inductor with a short circuit because we've assumed the SW has been held for a while.

So, we know that V1(t) = 6 V... I applied MESH ANALYSIS to the middle loop and to the right loop (both going clockwise) to find iL(0-).

Middle_Loop = 2A
Right_Loop = 2( iL(0-) - 2 ) + 12*iL(0-) = 30V... I get iL(0-) = 32/14 amps going CW

Is that right? If so, isn't it safe to assume that just after the switch is opened, for that one instant in time, the inductor will act like a current source since the current through an inductor CANNOT change instanteously.

So, I believe that you have to rewrite the circuit with the switch open and put a temporary current source in place of the inductor. After that, what do I do though? That's where I'm lost.
 
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Answers and Replies

  • #2
SGT

Homework Statement



http://img167.imageshack.us/img167/7747/picture2mg7.png [Broken]

Homework Equations



iL(t) = current through inductor

Using The Step by Step Method:

Assume the solution iL(t) = K1 + K2*e^(-t/tau)
where K1 = iL(infinity) and K2 = iL(0+) - iL(infinity)

so, Assume the solution has the form:

iL(t) = iL(infinity) + [iL(0+) - iL(infinity]*e^(-t/tau)

Now, what I did was I calculated iL(0-) first... I replaced the inductor with a short circuit because we've assumed the SW has been held for a while.

So, we know that V1(t) = 6 V... I applied MESH ANALYSIS to the middle loop and to the right loop (both going clockwise) to find iL(0-).

Middle_Loop = 2A
Right_Loop = 2( iL(0-) - 2 ) + 12*iL(0-) = 30V... I get iL(0-) = 32/14 amps going CW

Is that right? If so, isn't it safe to assume that just after the switch is opened, for that one instant in time, the inductor will act like a current source since the current through an inductor CANNOT change instanteously.

So, I believe that you have to rewrite the circuit with the switch open and put a temporary current source in place of the inductor. After that, what do I do though? That's where I'm lost.

The voltage on the 2 ohm resistor is V1. So, the voltage on the 12 ohm resistor is V1 + 5 V1 = 6 V1.
The current through this resistor is 6V1/12 = 0.5 V1.
The current through the 2 ohm resistor is V1/2 = 0.5 V1.
The sum of the 2 currents is 2 A. So, [tex]i_L(0^-) = 1 A[/tex]
For T >0 there are 2 sources: the independent currnt source 2A and the controlled voltage source 5V1. The current [tex]i_L(0^+) = i_L(0^-) =1 A [/tex]is only an initial condition and is not taken in consideration in forming the differential equation.
 
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  • #3
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But for the middle 2k resistor, I get 3Amps. and for the iL, I get 6V1/12 = 3 amps... how does this make sense?
 
  • #4
SGT
But for the middle 2k resistor, I get 3Amps. and for the iL, I get 6V1/12 = 3 amps... how does this make sense?
It makes sense when the switch is closed.
[tex]i_L(0^-) = i_L(0^+) = 6A[/tex].
The calculation I made before is valid for [tex]i_L(infinity)[/tex]
 
  • #5
180
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I got the ans, thx.
 

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