Question

An ac source of 50v,100÷πHz is connected in
series with an inductor of inductance L ad a
resistor of resistance R.the current in the circuit
is 2A ad p.d. across L and R are 30v ad 40v
respectively
1) calculate the average power dissipated ?
(2) calculate d power factor in the circuit?

Answer 1

average power: VI=40*2 watts
power factor

Vl and Vr are 90 degrees out of phase, so power factor=40/50=.8

Power factor is real power dissipated ratio to rms current times voltage of the circuit.

Answer 2

well, maybe V = 50 sin (2pift)
and i = I sin (2pift - phi)
then di/dt = 2pif I cos (2pift-phi)
with your numbers
2 pi f = 200
V = 50 sin 200 t
i = 20 sin (200 t-phi)
di/dt = 4000cos (200 t - phi)
now
VL = L di/dt
= L*4000 cos(200t-phi)
so 4000 L = 30
L = 3/400
Vr = iR =R*20 sin(200t-phi)
so 20 R = 40
R = 2 ohms

V = VL + VR
50 sin 200t = 30cos(200t-phi)+40sin(200t-phi)

30 cos(200t-phi) = 30cos200tcosphi+30sin200tsinphi

40sin(200t-phi) = 40sin200tcosphi-40cos200tsinphi

so
50 = 30 sin phi + 40 cos phi
and
00 = 30 cos phi - 40 sin phi

tan phi = .75

phi = 36.9 deg
multiply by pi/180 to get radians

sure enough 30*sin36.9+40*cos36.9 = 50 :)

so now I can do anything
V = 50 sin (200 t)
i = 20 sin (200 t-phi)

average power = average of V*i

power dissipated in R is
average of Vr*i
which is i^2 R
= 2^2 amp^2 * 2 ohms /2
we divide by 2 because average value of sin^2 or cos^2 is 1/2
= 4 watts