Test: AC Power Analysis- 2

Here, time period is, T = 20 ms; between 10 ms < t < 20 ms, i(t) = 0 A.


Also, 

= 50 x 0.01 =0.5 A

∴ P_avg = I² R
Now,

(since v and i in phase)

∴ instantaneous power,

Since V₁ < V₂, therefore the current wil! flow from V₂ to V₁ as shown below.


Hence, the whole power in the circuit will be supplied by V₂ only. Total power delivered by the voltage source V₂ is
P_T = I² x (2 + 4) + 6I
= 3² x 6 + 6 x 3 = 54 + 18
= 72 watts

V_R = (10 + 5 V_R) x R
Given, V_R = 5 V
So, 5 = (10 + 5 x 5) R
or, 

∴ Power loss in R is

= 54.30 A

V = (100 - j50) volt,
I = (4 - j3) A
∴ S = VI* = (100 - j50)(4 - j3)*
= (100 - j50)(4 + j3)
= 550 + j100
∴ P = 100 W and Q = 100 VAR
Q is in VAR not in VA. Hence, assertion is not correct.
Also P is the real part of apparent power (S = VI).

Instantaneous power in a purely resistive circuit is

In a purely inductive circuit, power consumed = Q = VI sin φ. In inductive circuit (having both R and L) power is consumed by resistor R which is P = VI cos φ.

Phase difference between v and i is

∴ power factor = cos φ = cos 75° lag
Since, i lags v therefore, the circuit elements may be resistance and inductance (Z = R + jX_L) or it may be resistance, inductance and capacitance Z = R + j(X_L - X_C), so that p.f. becomes lagging.