Electronics - Have I found the transient current correctly?

[DanRow93]

Homework Statement

A step voltage of 120v is applied to a series CR circuit. C = (4x10^(-6))F R = (20x10^(3))Ω
When the capacitor is fully charged the supply voltage is removed. Using KVL and Laplace transforms obtain an equation for the transient circuit current, given when t = 0, the voltage across the capacitor is at its steady state value of 120v.

Homework Equations

The Attempt at a Solution

V = v_r + v_c

0 = Rdq/dt + q/c

Laplace Transform 10:

0 = R(sq(s) - q_0) + (q(s))/c

q_0 = VC = 120 x (4x10^-6) = (4.8x10^(-4))
I'm not sure about this part, I think q_0 should be this.

0 = (20x10^3) (sq(s) - (4.8x10^(-4)) + (q(s)) / (4x10^(-6))

Divide through by R = (20x10^3):

0 = sq(s) - (4.8x10^(-4)) + (q(s)) / ((4x10^(-6)) (20x10^3))

Rearranging this gave:

q(s) = (4.8x10^(-4)) / (s+(25/2))

= (4.8x10^(-4)) x (1/(s + (25/2)))

Laplace Transform 2:

q = (4.8x10^(-4)) e^(-25t/2)

i = dq/dt = d/dt ((4.8x10^(-4)) e^(-25t/2))

i = -(3/500) e^(-25t/2) A

Does this seem right? I have a negative answer for current, and it is the same as the transient I found in an earlier question where I had to find the transient current when the voltage was applied, but this time negative.

Any help on where I've gone wrong is appreciated! If possible could you use the differential equations like I have. Thanks!

 

[vela]

It would help if you didn't plug numbers in until the end. I don't see any obvious mistakes. Is there a reason you think it's wrong?

Symbolically, you should get $q(t) = q_0 e^{-t/RC}$. If you differentiate this expression, you should find that $i(0) = -V_0/R$. The negative sign should make sense to you, and why the magnitude is $V_0/R$ should as well.

 

[gneill]

Does this seem right? I have a negative answer for current, and it is the same as the transient I found in an earlier question where I had to find the transient current when the voltage was applied, but this time negative.

A negative value for the current just means that your choice of direction for the current definition happened to be contrary to what actually is going on. You made that choice perhaps unknowingly when you wrote the equations. More specifically, dq/dt being the change in charge on the capacitor happens to actually be negative when the capacitor is discharging, whereas the math was written assuming current flowing into the capacitor and dq positive. No worries, just interpret the current direction appropriately.

 

[DanRow93]

It would help if you didn't plug numbers in until the end. I don't see any obvious mistakes. Is there a reason you think it's wrong?

Symbolically, you should get $q(t) = q_0 e^{-t/RC}$. If you differentiate this expression, you should find that $i(0) = -V_0/R$. The negative sign should make sense to you, and why the magnitude is $V_0/R$ should as well.

I wasn't quite sure about q_0 = VC ; if I should use 120v or 0v as the voltage supply has been switched off.

 

[DanRow93]

A negative value for the current just means that your choice of direction for the current definition happened to be contrary to what actually is going on. You made that choice perhaps unknowingly when you wrote the equations. More specifically, dq/dt being the change in charge on the capacitor happens to actually be negative when the capacitor is discharging, whereas the math was written assuming current flowing into the capacitor and dq positive. No worries, just interpret the current direction appropriately.

So to clarify, it makes sense that the current flows in the opposite direction around the circuit when the supply voltage is switched off and the capacitor is discharging, when compared to the current direction when the supply voltage is switched on?

 

[gneill]

So to clarify, it makes sense that the current flows in the opposite direction around the circuit when the supply voltage is switched off and the capacitor is discharging, when compared to the current direction when the supply voltage is switched on?

Yes. The capacitor is then supplying the emf that drives the current, and it will be in the opposite direction from the current that initially charged the capacitor.

 

[DanRow93]

Yes. The capacitor is then supplying the emf that drives the current, and it will be in the opposite direction from the current that initially charged the capacitor.

Great, thanks!

 

[vela]

I wasn't quite sure about q_0 = VC ; if I should use 120v or 0v as the voltage supply has been switched off.

$Q = CV$ is the defining relationship for a capacitor. Just remember that $V$ corresponds to the voltage across the capacitor. When the voltage supply is set to 0 V, that has nothing to do with the voltage across capacitor, which still has the charge it had the instant before.