Question

I just need explanation:

Question: In the interval 0 12 years ago

Answer 1

It says on the book that:
\[v=L \frac{ di }{ dt } =15.0\cos50t (V)\] \[p=vi=75.0\sin 100t (W)\] \[w _{l}=\int\limits_{t}^{0}pdt=0.75(1-\cos100t)(J)\]

Answer 2

\[v = L \frac{ di }{ dt } = 30.10^{-3} . \frac{ d(10\sin50t) }{ dt } = 30.10^{-3} .10.50\cos(50t)=15\cos(50t)\]

Answer 3

\[p = vi = 15\cos50t . 10\sin50t = 150.0.5.\sin100t = 75\sin100t (W)\]

Answer 4

so sin50*cos50 is sine 100?

Answer 5

\[w_t = \int\limits_{0}^{t}pdt = \int\limits_{0}^{t}75\sin(100t)dt = \frac{ 75.(-\cos100t) }{ 100 } - \frac{ 75.(-\cos0) }{ 100 } = 0.75(1-\cos100t)\] (J)

Hint: sin2x = 2sinx.cosx.
One note that, the formula for energy should integrate from 0 --> t, not t --> 0 as your book said

Answer 6

your post is cut

Answer 7

Yes, you can solve the rest by yourself, it cut due to limit of page :D, maybe a good way to try to learn from the rest

Answer 8

yes,,thanks

Answer 9

btw..it is only sin50*cos50 not 2sin50*cos50

Answer 10

No, you can see my answer, I put 0.5 factor, which means, sin50t*cos50t = 0.5sin100t, which means similar to sin100t = 2sin50t*cos50t

Answer 11

without 0.5, you can not get the correct answer ;) Cheers

Answer 12

ohh....ok thank you