Question

In the circuit given below, determine the mesh currents i1(t) and i2(t). Let v1 = 14 cos(4t) V and v2 = 20 cos(4t – 30°) V. Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 180 degrees.

i₁(t) =  cos(4t + (˚)) A

i₂(t) =  cos(4t + ˚) A

Answer 1

frequency omega = 4 rad/sec Inductive reactance of 1H inductor is X_L = omega L = 4 times 1 = 4 ohm Capacitance reactance of 1F capacitor is X_C = 1/omega C = 1/4 times 1 = 0.25 ohm Applying KVL in loop (1) we get (1 + j4)i_1 + (1 - j0.25) (i_1 - i_2) = 14 rightangle 0 degree (2 + 3.75j)i_1 - (1 - j0.25)i_2 = 14 rightangle 0 degree rightangle (1) Applying KVL in loop (2) we get (1 + j4)i_2 + (1 - j0.25) (i_2 - i_1) = -[20 rightangle -30 degree] - (1 - j0.25)i_+ 1 + (2 + 3.75j)i_2 = -(20 rightangle -30 degree) rightarrow (2) i_1 = 20 rightangle -30 degree + (2 + 3.75j)i_2/(1 - j0.25) \r\n Substitute i_1 value in equation (1) we get (2 + 3.75j) [20 rightangle -30 degree + (2 + 3.75j)i_2/(1 - j-.25)] - (1 - jo.25)i_2 = 14 rightangle 0 degree 57.32 + 59.29 = 8j + (-13 + 11.75 j)i_2 - (1 - j0.25)i_2 = 14 rightangle 0 degree

Answer 2

1-S2 1H T1P 14 cas (4t) 1 2 14-30 20 Les Drau the ckt in phasor domain tregen have same 4 1 S 20 L-30 Di 14 L0 APply kVL rop + 14 LO t - 2 14 LO 0 1- in dupp KVL Appy + 을스90 + 20 L-30

Dn Solning eqn aud 2-8 194.44 4irms 2 534 L46-52 irms 17 time domain Cos eot+o) Peak 2.5342 cos (4t-46.52) l) 2081 2 cos 4t944y 2It) i, Lt)30 58 Cos 14t-46.52) 1)3.97 cos 14tt 34.44)