1.Vs=25 cos(40t + 45).
Vs=25 sin(90 – 40t -45)
since cos(a)=sin(90-a).
Vs =25 sin(45 -40t)
Vs= 25 sin(180 -45 +40t)
as sin(a)=sin(180-a).
So , Vs=25 sin(40t +135)
Vs =25 ∠1350.
The given figure has two loops :- one on the left and the other on the
right.
First, the impedance of an inductance is given by Xl = j*w*L, where
w is the angular frequency and L is the value of inductance.
In this case , value of angular frequency w=40
To find i1 and i2, we have to apply Kirchoff’s Voltage law
1.Applying KVL in the first loop ,
The first loop has a voltage source, inductance of 8H and a
resistance of 400 Ohm. Impedence of inductance is Xl =j * 8 * 40
Xl =320j
So , by KVL
Vs=320j * i1 + 400(i1 – i2)
(1)
2.Applying KVL in the second loop,Here the inductance add in series
and combined inductance is 9H. And combined Inductive reactance
is Xl =j*40*9 or Xl=360j.
400(i1 –i2) = 50 * i2 +360j*i2
400*i1 =450 *i2 + 360j*i2 (2)
Solving equations (1) and (2),
Vs = i2 *(720j -238)
Or
25 ∠1350. = i2 * 758.31 ∠108.290
i2 = 32.96 ∠26.710 mA,
(3)
From equation (2) we see that
i1 = 1.44 ∠38.650 * i2
From (3) and (4)
i1 = 47.46 ∠65.360 mA
(4)