Given I, find V across the elements (resistor, inductor, and capacitor)

[courteous]

Homework Statement

The current after $$t=0$$ in a single circuit element is as shown in Fig. 2-37. Find the voltage across the element at $$t=6.5\mu s$$, if the element is (a) $$10 k\Omega$$, (b) $$15 mH$$, (c) $$0.3 nF$$ with $$Q(0)=0$$.

Ans. (a) 25 V; (b) -75 V; (c) 81.3 V

[PLAIN]http://img684.imageshack.us/img684/2684/unled1copyb.jpg

Homework Equations

(a) $$V=RI$$

(b) $$v(t)=L\frac{di(t)}{dt}$$

(c) $$v(t)=\frac{1}{C}\int i(t)dt$$

The Attempt at a Solution

I'm having problem with the (c) case.

(a) $$V=10\text{ }k\Omega \times 2.5\text{ }mA=25\text{ }V$$

(b) $$v(t)=15\frac{(0-5)mA}{(7-6)\mu s}=-75\text{ }V$$

Now, what are the limits of integration for capacitor? Do you integrate from $$t=0$$ to $$t=6.5\text{ }\mu s$$? If so, why?

I did use this domain, but the result is incorrect:
(c) $$v(t)=\frac{1}{0.3\text{ }nF}\int_0^{6.5} i(t)dt=\frac{1}{0.3\text{ }nF}\left(\frac{1}{2}(3\text{ }\mu s\times 5\text{ }mA)+3\text{ }\mu s\times 5\text{ }mA+\frac{3}{4}(5\text{ }mA\times 1\text{ }\mu s)\right)=87.50\text{ }V\neq 81.3\text{ }V$$

 

[KataKoniK]

Hi there,

For the capacitor, you would need to integrate from t=0 to t=6.5 microseconds because that is the time frame specified in the problem. The voltage across the capacitor at any given time is determined by the current flowing through it at that time, so you need to take into account the entire time frame given in order to get the correct voltage.

In your solution, it looks like you may have made a mistake in the calculation. The correct solution should be:

(c) v(t)=\frac{1}{0.3\text{ }nF}\int_0^{6.5\text{ }\mu s} i(t)dt=\frac{1}{0.3\text{ }nF}\left(\frac{1}{2}(3\text{ }\mu s\times 5\text{ }mA)+3\text{ }\mu s\times 5\text{ }mA+\frac{3}{4}(5\text{ }mA\times 1\text{ }\mu s)\right)=81.25\text{ }V

This would give you the correct answer of 81.3 V, as stated in the forum post. I hope this helps!