# Circuit Theory Problem - Current and Charge

1. Aug 25, 2014

### aflyingcougar

1. The problem statement, all variables and given/known data
P 1.2-4

2. Relevant equations
I know that i(t) [current w/respect to time] = dq/dt. I also know that by the previous equation, q(t) = ∫i(τ) from -∞ to t (sec).

3. The attempt at a solution
I understand simpler problems involving finding q(t) by using the integral relationship with i(t), but for this problem, I am confused on how to set my bounds for integration.

My guess would be to integrate i(t) from t = 2 to 8 because that is the domain in which i doesn't equal zero. However, based on the answer (bottom of picture) I see the answer is in terms of t seconds. This is what is confusing me. I don't see how to solve this without just getting an integer. So, I guess my problem is that I don't know how to set up the integration.

Can someone help explain to me how you would decide to set up the integration to find q(t) ? Thanks..I really am trying to understand this.

2. Aug 26, 2014

### jz92wjaz

For this problem, it's an indefinite integral.

For example, i(t) for 2 < t < 4 is 2
q(t) = ∫i(t)dt = ∫2dt = 2t + C for 2 < t < 4

Evaluate at q(2) to find C.

Last edited: Aug 26, 2014
3. Aug 26, 2014

### Simon Bridge

As written: it is asking for the total charge that has entered the circuit element after t=0 - not the instantaneous charge entering at some time t. But that is not what the answer seems to be showing.

I'd do this as a series of DEs ... i.e. for
0<t<2 dq1/dt = 0 so q1=?
2<t<4 dq2/dt = 2 so q2=?

As jz92wjaz you can do this as an indefinite integral and use the boundary conditions to find the values of the constants of integration. i.e. does q1(t=2)=q2(t=2) ??

You could probably also do this by sketching the i(t) graph and finding the areas geometrically.

4. Aug 26, 2014

### aflyingcougar

Thank you so much. This makes sense. It's been a while since I've practiced calculus. There is one more thing I don't understand. How and why would I pick to evaluate q(2) to find the c in q2 = 2t + c. If the bounds on the piece wise function remain the same, it appears that q(2) doesn't exist.

Edit: also I don't know what to pick as the initial condition. Why pick 2?

Last edited: Aug 26, 2014
5. Aug 26, 2014

### Staff: Mentor

Whatever way you solve this, you should first sketch the graph of accumulated charge vs. time. It's a couple of straight line segments.

Then just figure out on the sketch the equations of those straight lines (using the slope and the projected y-intercept). No calculus needed!

As a supplementary exercise, once you have it worked out the easy way, go back and do it using calculus.

6. Aug 26, 2014

### aflyingcougar

Thanks Nascent. I see how I can solve it in that manner, which is much easier. However, I think my ELC professor wants to see the work via calculus.

7. Aug 26, 2014

### Simon Bridge

The sketch everyone wants you to do will explain how and why you evaluate the constants of integration.
If you look at the answer, you'll see that the intervals have the same value if evaluated at the boundary time. i.e. the 1st and 2nd intervals at t=2 are both 0. That should be a hint.

I know - the peicewise function is written with no value at t=2 (etc) ... that does not mean that the charge does not exist there, only that the author of the question was being sloppy.

8. Aug 26, 2014

### Staff: Mentor

When integrating, the initial condition is the charge on the capacitor at the start of that charging period, i.e., when you change to the new value of current.

9. Aug 26, 2014

### aflyingcougar

I drew the graph, and I see how you can extend the line to figure out the y-intercept, thus an initial value. Thank you for that. So, re reading everything, I have figured out many different way to solve this. Would the differential equation approach to this (from post #2) be the only way to solve this without having to reference the graph of q(t)?

10. Aug 27, 2014

### Simon Bridge

I think any other approach requires some measure of prescience (read: experience with this sort of problem.)

I think you can set up the integral for cumulative charging as: $$q(t)=\int_0^t i(t')\;dt'$$ ... but you are best doing this sort of problem by thinking about the physics rather than the mathematics.

11. Aug 27, 2014

### Staff: Mentor

I wouldn't make a distinction. Even were the current to be a complicated function of time, you should still sketch it as a guide before using maths to get the precision needed. Whether you perform that maths in your head or on paper really doesn't make it a different approach.

12. Aug 27, 2014

### aflyingcougar

Alright. I understand. Thanks everyone!