Question

Problem 4 The current l(t)-3A sin (80t/s) flows through a 15Ω resistor. (i) What is the frequency of the current? (ii) What is the rms value of the current? (ii) Determine the equation that describes the potential across the resistor.

Answer 1

Given equation for current I(t) = 3 A sin (80 t/s) & R = 15 Ohm. I(t) = 3A sin (80 t/s) is given in term of frequency (f), not in terms of w: where (w = 2 pi f) So: from equation frequency [f = 80 sec^-1] \r\n Given peak value of current I_0 = 3 Amp. Therefore I_rms = I_0/squareroot 2 = 3/squareroot 2 Amp = 2.12 Amp Potential across the resistor R is given by v(t) = I(t) R = 3 times 15 A sin (80 t/s) v(t) = 30 A sin (80 t/s)