00:01
For this problem on the topic of ac circuits, we have a sinusoidal ac voltage of 40 volts rms and a frequency of 100 hertz being applied to firstly a 100 -oom resistor, then a 0 .2 -henductor, and lastly a 50 micro -faird capacitor.
00:18
In each case, we want to find the peak value of the current as well as the average power delivered.
00:24
Now for part a, the peak current across the resistor, or the current as the function of time rather i r of t is equal to the peak current which is v0 over r times sine of omega t and so the peak current across the resistor i r not is v0 over r which is the rmsv voltage times the square of two over the resistance r these values are known so putting them in we get this to be 40 volts times the square root of 2 divided by the resistance 100 oms.
01:09
This gives the peak current of 0 .6 amperes.
01:17
The power delivered in this case across the resistor, the average power is a half times v0 squared over r, which is equal to a half times the rr, rms voltage squared times the square root of 2 squared over r.
01:49
And so this is the rms voltage squared over r, which is 40 volts squared over the resistance 100 oms and we get an average power delivered to the resistor to be 16 watts or to one significant figure two times 10 to the power 1 watts.
02:24
Now for part b we want the peak current through the inductor.
02:28
We know the current through the inductor as a function of time is v0 over omega l times the cosine of omega -t...