Power will not be equal to 0.
P = V^2/R = I^2 R which are never negative.
Mean power = Energy in 1 cycle / Time for 1 cycle
Use the time base and y-gain.
Energy dissipated in 1 cycle = Energy during 0 to 1 ms + Energy during 1 to 1.5 ms because first 1.5ms is one cycle
so ans to (b)(iii) = ans to (b)(i) + ans to (b)(ii)
Edited to explain mean power = energy dissipated in 1 cycle / time for 1 cycle:Power dissipated in 1 cycle = energy dissipated in time for 1 cycle / time for 1 cycle = energy dissipated in T seconds / T
Mean power dissipation = energy dissipated in T seconds / Time for 1 cycle = energy dissipated in T seconds / T
ans to (b)(iv) = energy dissipated in 1 cycle / time for 1 cycle = ans to (b)(iii) / time for 1 cycle = ans to (b)(iii) / period of signal = ans to (b)(iii) x frequency of signal = ans to (b)(iii) x ans to (a)(iii)
Average velocity,
Unparseable latex formula: v \bar
=
t2−t11∫t1t2v(x)dxe.g
v(t)=2Then average velocity for t = 3 to t = 5,
v=5−31∫352dx=2If motion is periodic and it is meaningful to talk of cycles, then v = displacement over 1 cycle / time for 1 cycle
I do not understand "velocity occur for 1 cycle".