Energy dissipated in a resistor during a time interval?

[mkematt96]

Homework Statement

Homework Equations

P = I*V energy dissipated = wr = ∫ ( P *dt) ... t is from 0 to .4 seconds vs = 400t^2 = 400 * (.4^2) = 64 V

The Attempt at a Solution

Using KVL I said Vs ( 64 V) = 100 * i ... I found that I was 64/100 = .64 amps. I then said power is .64 amps ^ 2 * 100 ohms = 40.96 Watts. I then integrated 40.96 Watts with respects to time ... Energy dissipated = wr = ∫ ( 40.96 Watts *dt ) from 0 to .4 seconds = 40.96 * .4 = 16.384 J.Where did I go wrong?

 

[BvU]

The 40.96 is not the power at $t<0.4$ s. You calculate as if the voltage was a constant 64 V from $t= 0$ to $t=0.4$ s.

 

[mkematt96]

The 40.96 is not the power at $t<0.4$ s. You calculate as if the voltage was a constant 64 V from $t= 0$ to $t=0.4$ s.

So then if the voltage is changing I could integrate with respect to time: Vs = ∫ 400t^2 dt from 0 to .4 s which is equal to 8.533 V then divide by 100 ohms to find the current .08533 amps. Then integrate power with respect to time: Wr= ∫ ( .08533^2 * 100) dt from 0 to .4 sec ?

 

[BvU]

Aren't you doing something similar again ? Why integrate twice ? Use your own relevant equation $\displaystyle \int_0^{0.4 {\rm s}}\; P(t) \;dt$ where you substitute an appropriate expression for $P(t)$