# Help with an Inductor question with voltage vs time graph

Discussion in 'Homework Help' started by gengm, Apr 7, 2010.

1. ### gengm Thread Starter New Member

Apr 7, 2010
23
0
Hello everyone, new to the boards here and I'm looking for some help with an Inductor question.
Here is the question.

I'm given the equation for power stored, W = (1/2)Li^2
In which i is a function of the voltage in which i = 1/L * integral of V*dt
1st off I switched the L = .5H, then at given time 100ms, the Voltage is 30V

So I did i = 30V/.5H = 60 for the current
then inputed it into the W equation and got .5*.5*(60^2) = 900J
but 900 isn't the correct answer.
Can someone point out where I went wrong?

2. ### Ghar Active Member

Mar 8, 2010
655
72
Where did you get i = 30V/0.5H ?
What happened to the integral?

3. ### gengm Thread Starter New Member

Apr 7, 2010
23
0
Considering they want the energy stored at a certain time, wouldn't the voltage at that time be the only thing relevant?
then from that equation i = (1/L)*V

4. ### Ghar Active Member

Mar 8, 2010
655
72
That equation refers to current and you don't know the current at any time until you calculate it.
Current is the integral of voltage, meaning voltage for all time up to the time of interest matters.

Specifically, you get:

$
i = \frac{1}{L} \int_0^{40ms} 80 dt \quad+\quad \frac{1}{L}\int_{40ms}^{100ms}30 dt
$

Last edited: Apr 7, 2010
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5. ### gengm Thread Starter New Member

Apr 7, 2010
23
0
Alright, that's where I got mixed up there. Assumed that at that time 100ms, thought that only mattered.

Much thanks Ghar.

Mar 8, 2010
655
72
No problem