Electrical engineering question?

a sinusoidal current, having a maximum value of 12 A, flows through a circuit of 10 ohms

calculate the p.d across the resistor
calculate the power dissipated in the resistor

Answer:
PD=IR=12*10=120 V
Power=I^2 * R = 144*10=1440 W

For a sinusoidal current root mean square (RMS) value is Irms=Ipeak/sqrt(2)=12/1.41=8.5

p.d across the resistor =Irms*R=8.5*10=85V
power dissipated in the resistor=Irms^2*R=722.5W

First you have to calculate the effective value of the sinusoidal alternating current with a peak value of 12A

I-eff = I-peak / sqrt2 = 12 / 1.4146

P = I^2 * R = (12 / 1.4146)^2 * 10 = 72 * 10 = 720W

V = I-eff * R = 12 / 1.4146 * 10 = 8.485A * 10 Ohm = 84.85V

Use the RMS value (also called the 'effective' value) of the current.

12 / sqrt(2) = 8.485 amps RMS

Power = I^2 * R = 8.486^2 * 10 = 720.0 Watts.

note: the formula "divide by the squareroot of 2" only works for sinusoidal waves. For other waveshapes, there are different formulas for calculating RMS values.

.

P = V * I = R * I ^ 2 = V^2 / R = 10 * (12cos(wt + k)) ^ 2 = 10 * 144*(1/2 + cos 2(wt + k)/2) ______ this is answer in time domain

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