Alternating Current 6: Period, Load Resistance, RMS & Peak Power

[islandguy]

6. A pure alternating current is produced by an alternator in which a coil rotates in a uniform magnetic field. The output voltage from the alternator can be described by V=12sin100t.
a) What is the period of the alternating voltage produced?
[ marks]
b) The output from the alternator is connected across a resistive Load of 6 ohms.
i) Write down the equation that describes the variation of current
through the load.
[ mark]
ii) Calculate the r.m.s value of the current.
[ mark]
iii) Calculate the peak value of the power consumed by the load.
[ mark]
iv) Sketch a graph of power developed across the resistor, as time passes.


Thanks for ur help

 

[Locrian]

This is a plug-n-chug type of question. What kind of problem are you having with it?

 

[M98Ranger]

a) The period of the alternating voltage produced is 1/100 seconds or 0.01 seconds.

b) i) The equation that describes the variation of current through the load is I=V/R, where V is the voltage and R is the resistance. In this case, the voltage is given by V=12sin100t, so the current can be written as I=(12sin100t)/6.

ii) To calculate the r.m.s value of the current, we need to square the current equation and then take the average over one period. This can be written as:

Irms = √(∫(I^2)dt/T)

= √(∫((12sin100t)/6)^2dt/T)

= √(∫(12^2sin^2(100t))/36dt/T)

= √((12^2/36)∫sin^2(100t)dt/T)

= √((12^2/36)(T/2))

= √((12^2/36)(1/200))

= √(12^2/7200)

= √(144/7200)

= 0.2 amps

Therefore, the r.m.s value of the current is 0.2 amps.

iii) To calculate the peak value of the power consumed by the load, we need to use the equation P=I^2R, where I is the r.m.s current calculated in part (ii) and R is the resistance. In this case, the peak power can be written as:

Ppeak = (0.2)^2 x 6 = 0.24 watts

Therefore, the peak value of the power consumed by the load is 0.24 watts.

iv) The graph of power developed across the resistor as time passes would look like a sine wave, with the peak power occurring at the peaks of the sine wave and the lowest power occurring at the zero crossings. The x-axis would represent time (in seconds) and the y-axis would represent power (in watts). The graph would look something like this:

[Graph of power developed across the resistor, with time on the x-axis and power on the y-axis, showing a sine wave with peaks at the top and bottom and zero crossings in between]