Hello,

[see corrections later in this thread]

Im not sure how much progress you have made so far, but just to try to help a little more here...

You can reduce the circuit to a source and impedance. This might make it easier for you to understand.
If you redraw the circuit to position the current source where the capacitor is now and the capacitor where the current source is now you'll see you just have an impedance being driven by a current source. You can reduce the three elements L,C, and R, down to one impedance. The output across the resistor is then the current times this impedance.

To combine into one impedance you can add the impedance of the cap to the inductors, then put that in parallel with the resistor. The output is then Vout=I*Z where Vout is the voltage across the resistor.

In many circuits that have both an L and a C the response could be a second order damped sinusoidal response. Double check for this using a circuit simulator, and i think you will find there should be a sinusoidal term in the solution as well as an exponential. So something like A+B*e^(-at)*sin(wt) (where B could be equal to A) might be expected.

 

Hello,

Im not sure how much progress you have made so far, but just to try to help a little more here...

You can reduce the circuit to a source and impedance. This might make it easier for you to understand.
If you redraw the circuit to position the current source where the capacitor is now and the capacitor where the current source is now you'll see you just have an impedance being driven by a current source. You can reduce the three elements L,C, and R, down to one impedance. The output across the resistor is then the current times this impedance.

To combine into one impedance you can add the impedance of the cap to the inductors, then put that in parallel with the resistor. The output is then Vout=I*Z where Vout is the voltage across the resistor.

In many circuits that have both an L and a C the response could be a second order damped sinusoidal response. Double check for this using a circuit simulator, and i think you will find there should be a sinusoidal term in the solution as well as an exponential. So something like A+B*e^(-at)*sin(wt) (where B could be equal to A) might be expected.

This won't work because the voltage being asked for is NOT the voltage across the effective combination of all three elements. Think about it, using your approach you would have a current source driving an impedance. Well, that means that the current going into that impedance is simply the current output of the source.

As for the sinusoidal term, that depends on the damping factor. In this case, the circuit is critically damped, so there is no sinusoidal term.

 

This won't work because the voltage being asked for is NOT the voltage across the effective combination of all three elements. Think about it, using your approach you would have a current source driving an impedance. Well, that means that the current going into that impedance is simply the current output of the source.

As for the sinusoidal term, that depends on the damping factor. In this case, the circuit is critically damped, so there is no sinusoidal term.

Hello again,

Yes, you are absolutely correct there. I looked at the circuit a little too fast this time and determined it to be a little different. It was nice of you to point this out.
We can still use impedance combinations however so i'll explain that a little.

Again swap the positions of the current source and capacitor to make the drawing a little more clear.
Next, we can combine L and R impedances to make one impedance Z1, then we can combine Z1 with the impedance for C, call that Z2.
Now the voltage at the left side of the inductor is I*Z2, and the output voltage is then I*Z2*R/Z1.
So in other words, we can combine impedances for L and R in series, then parallel that with impedance for C, then calculate the voltage at the top of C and left side of L, then calculate the output voltage across R.
The result is then (I*R)/(s*(s^2*C*L+s*C*R+1))
which when the values are plugged in and the transform taken, we get:
v(t)=20-40*t*e^(-2*t)-20*e^(-2*t)

This happened to come out as critically damped, however this is very rare in real life.

 

Hello again,

Yes, you are absolutely correct there. I looked at the circuit a little too fast this time and determined it to be a little different. It was nice of you to point this out.
We can still use impedance combinations however so i'll explain that a little.

Again swap the positions of the current source and capacitor to make the drawing a little more clear.
Next, we can combine L and R impedances to make one impedance Z1, then we can combine Z1 with the impedance for C, call that Z2.
Now the voltage at the left side of the inductor is I*Z2, and the output voltage is then I*Z2*R/Z1.
So in other words, we can combine impedances for L and R in series, then parallel that with impedance for C, then calculate the voltage at the top of C and left side of L, then calculate the output voltage across R.
The result is then (I*R)/(s*(s^2*C*L+s*C*R+1))
which when the values are plugged in and the transform taken, we get:
v(t)=20-40*t*e^(-2*t)-20*e^(-2*t)

This happened to come out as critically damped, however this is very rare in real life.

nice idea , I will do same this
thanks sir