It may be kept in mind that the coil has not only inductance L but also some resistance r which produces power loss. In the voltage vector diagram, AB represents drop across R = 25 V. Vector BC represents drop across coil which is due to L and r. Which value is 40 V and the vector BC is at any angle of φ with the current vector. AD represents 50 V which is the drop across R and coil combined. AE represents the drop across the capacitor and leads the current by 90°.
It will be seen that the total horizontal drop in the circuit is AC and the vertical drop is AG. Their vector sum AF represents the applied voltage V.
From triangle ABD, we get 502 = 402 + 252 + 2 × 25 × 40 × cos φ ∴ cos φ = 0.1375 and sin φ = 0.99.
Considering the coil, IZL = 40
∴ ZL = 40/0.345 = 115.94 Ω
Now r = ZL cos φ = 115.94 × 0.1375 = 15.94 Ω
Power loss in the coil = I2r = 0.3452 × 15.94 = 1.9 W
BC = BD cos φ = 40 × 0.1375 = 5.5 V; CD = BD sin φ = 40 × 0.99 = 39.6 V
AC = 25 + 5.5 = 30.5 V; AG = AE −DC = 55 − 39.6 = 15.4 V
AF = √(AC2 + CF2) = √(30.52 + 15.42) = 34.2V
