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Given the first order circuit above. The switch has been closed for a long time and is opened at \$t=0\$. Find the equation for the voltage \$vc(t)\$ across the capacitor after the switch has been opened.

I have determined that \$vc(0)=7.619 V\$, but cannot find the time constant.

Here is my work for attempting to get \$R_{th}\$.

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hint - what is the impedance of a current source? What does that contribute to the total impedance across the capacitor? –  WhatRoughBeast May 18 at 0:42
    
Does an ideal current source have an impedance? –  Chris Crutchfield May 18 at 0:43
    
What is the V/I relationship for an ideal current source? –  Zuofu May 18 at 0:49
    
I see that an ideal current source is supposed to have an infinite impedance, but I'm still not sure how that helps. –  Chris Crutchfield May 18 at 0:56
    
What is the parallel resistance of a resistor R and an infinite impedance? And how would this affect a time constant or a resistor and a capacitor (RC)? –  WhatRoughBeast May 18 at 2:04

2 Answers 2

This is slightly tricky given the presence of the controlled source.

While I won't work the problem for you, I will tell you that you can combine the controlled current source and the two rightmost resistors into one equivalent resistance \$R_{EQ}\$ which is then in parallel with the \$50k\Omega\$ resistor.

Here's a hint: the voltage across the two rightmost resistors (and thus, the current source) is just \$(15k\Omega + 25k\Omega)i(t) \$

But the current through the top wire connecting the capacitor to the current source / resistors is just \$0.75 i(t)\$. Thus, the equivalent resistance to the right of the capacitor is...

schematic

simulate this circuit – Schematic created using CircuitLab

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Would that mean that the test current should be .75i? Would that mean that R_th should be 53.333k ohms? –  Chris Crutchfield May 18 at 1:45
    
@ChrisCrutchfield, the equivalent resistance to the right of the capacitor is indeed 53.3k. We didn't use a test current - we just used KCL. However, we could have used a test current. Let the test current (to the right of the capacitor) be 1A. Then, the current through the resistors must be 4/3A in order to satisfy KCL. Thus, the voltage, due to a 1A test current, is (4/3)40k = 53.3kV which implies the equivalent resistance is 53.3k ohms –  Alfred Centauri May 18 at 1:56

Since this is a homework question, I will give an extended hint as an "answer". First, consider what the 'R' portion of the RC time constant means. This is the resistance which the capacitor must discharge through to fall to \$e^{-1}\$ of its original value. We know that the switch is open, so the left side of the circuit is effectively disconnected, we are concerned only with the right side of the circuit (to the right of the switch, including the 50 K resistor).

Normally, if we have only independent sources, we can remove them (by short-circuiting ideal V sources and open-circuiting ideal I sources), but this circuit has a dependent current source, so we cannot do that as easily. However, the circuit is linear, so we can still find the equivalent resistor from the perspective of the capacitor, which is what you have to do to solve for the time constant.

Recall the procedure to find the Thevenin resistance in a case with dependent sources. This is done by applying a test voltage (for example \$V_{test} = 1 V\$) at the terminals, and finding the resulting current (\$I_{test}\$). The equivalent resistance is then \$R_{Th}=\dfrac{V_{test}}{I_{test}}\$.

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Also note that you do not necessarily need to apply a test voltage and calculate the test current. Sometimes (as I think in this circuit) the math is easier to apply a test current and calculate the test voltage. In either case, as long as you have a V/I pair, you can calculate the equivalent resistance. –  Zuofu May 18 at 1:11
    
So would my R_th be 40k? –  Chris Crutchfield May 18 at 1:14
    
I think I made a mistake. Is R_th 22.222k ohms? –  Chris Crutchfield May 18 at 1:21
    
That's not what I got, upload your work and we can see. Also, ignore what I said about the test current - that's a good approach in general but I think it's a pain in this problem. –  Zuofu May 18 at 1:26
    
I edited my question with the work to find R_th. –  Chris Crutchfield May 18 at 1:37

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