00:01
In our question today we are given that the current entering the positive terminant of a device is given by equation 1 and the voltage across the device is given by equation 2.
00:11
We need to find the charge delivered to the device between the time interval of 0, 2 seconds.
00:21
Now the basic formula for current is given as dq by dt where charge q is given as integration of dt.
00:36
Applying the value of equation, we have charge equal to 2t, d, t.
00:48
Further simplifying, we have dt as a equation.
00:58
Performing the integration we get, now that we have a charge equation, at time, t being two seconds, the charge will be, e to the power minus 2 t 0 to 2 2 10 to the power minus 3 prime the conditions we get the charge equivalent to 2 .9 into 10 to the power minus 3 coolum respectively which we can write as 2 .9 0 .9 .000 coolum.
01:59
Next we are to determine the power absorb.
02:01
Now the power equation is given as of t multiplied by as per time which is the instantaneous power...