Post History
In a previous question we found that for a critically damped oscillation: $$ I(t) = 2 t e^{-0.5 t} $$But if we integrate to find the voltage through the capacitor: $$ V_{CO} + \frac{1}{C} \cd...
#3: Post edited
Voltage of tank circuit is invalid
In a [previous](https://electrical.codidact.com/posts/282585) question we found that for a critical damped oscillation:- But if we integrate to find the voltage through the capacitor:
the integral is always a positive value which doesnt make any sense since the capacitor is being discharged.What am I doing wrong here?Help appreciated!
- In a [previous](https://electrical.codidact.com/posts/282585) question we found that for a critically damped oscillation:
- $$
- I(t) = 2 t e^{-0.5 t}
- $$
- But if we integrate to find the voltage through the capacitor:
- $$
- V_{CO} + \frac{1}{C} \cdot \int_0^t{I(t)}dt
- $$
- the integral is always a positive value which doesn't make any sense since the capacitor is being discharged.
- What am I doing wrong here? Help appreciated!
#2: Post edited
- In a [previous](https://electrical.codidact.com/posts/282585) question we found that for a critical damped oscillation:
- 
- But if we integrate to find the voltage through the capacitor:
- the integral is always a positive value which doesnt make any sense since the capacitor is being discharged.
- What am I doing wrong here?Help appreciated!
- In a [previous](https://electrical.codidact.com/posts/282585) question we found that for a critical damped oscillation:
- 
- But if we integrate to find the voltage through the capacitor:
- 
- the integral is always a positive value which doesnt make any sense since the capacitor is being discharged.
- What am I doing wrong here?Help appreciated!
#1: Initial revision
Voltage of tank circuit is invalid
In a [previous](https://electrical.codidact.com/posts/282585) question we found that for a critical damped oscillation:  But if we integrate to find the voltage through the capacitor:  the integral is always a positive value which doesnt make any sense since the capacitor is being discharged. What am I doing wrong here?Help appreciated!