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
2007-06-28 00:50:43 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
PD=IR=12*10=120 V
Power=I^2 * R = 144*10=1440 W
2007-06-28 00:59:43 · answer #1 · answered by Azurri 2 · 0⤊ 1⤋
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
2007-06-28 01:43:05 · answer #2 · answered by Ernst S 5 · 0⤊ 0⤋
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.
.
2007-06-28 06:17:55 · answer #3 · answered by tlbs101 7 · 0⤊ 0⤋
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
2007-06-28 01:37:40 · answer #4 · answered by oleg_arch 2 · 0⤊ 0⤋
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
2007-06-28 10:11:20 · answer #5 · answered by mehdi 1 · 0⤊ 0⤋