Question

1. (i) In a circuit, the voltage is given by: Vs (t) 12 cos(50t10) V (a) What is the amplitude of the voltage? (b) What is the phase of the voltage? (c) What is the angular frequency (radians/sec.) of the voltage? (d) What is the frequency (in cycles/sec.) of the voltage? (e) Determine vs (t) at time t 2.5ms (ii) Given v1(t)cos(50t 50°) V and v2(t) 12 sin(50t 10°) V, determine the phase angle between the two sinusoids and determine which one lags the other (iii) Evaluate the following complex numbers and express the results in x + jy (rectangular) form and in rz0 (polar) form 3+j4 (a)2 + 5-j8 (b) 4L-100 )2 3260 82100+62-200 92800-42500 (iv) Using phasors, simplify the following into single sinusoids (a) 3cos(20t10°) 5cos (20t 30°) (b)20sin(400t) 10 cos(400t 60) 5sin(400t 20°) [5+5+5+5 Marks]

Answer 1

v_s (t) = 12 cos (50 t + 10 degree) v Amplitude 12 v Phase phi = 10 degree Angular frequency (w) w = 50 rad/s cycles cycles = (50/2 pi) cycles/sec = 7.96 cycles/sec v_s (t) = 12 cos (50 t + 10 degree) t = 2.5 m s = 2.5 times 10^-3 sec = 12 cos (50 times 2.5 times 10^-3 + 10) = 11.81 volt \r\n v_1 (t) = -cos (50 t + 50 degree) = -sin[pi/2 - (50 t + 50 degree)] = -sin(-50 t + 40 degree) v_1 (t) = sin (50 t - 40 degree) v_2 (t) = 12 sin(50 t - 10 degree) v_1 (t) lags v_2 (t) by 30^degreE Phase angle between v_1 & v_2 is 30 degree.