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Two impedances Z1 and Z2 when connected separately across a 230-V, 50-Hz supply consumed 100 W and 60 W at power factors of 0.5 lagging and 0.6 leading respectively. If these impedances are now connected in series across the same supply, find : (i) total power absorbed and overall p.f. (ii) the value of the impedance to be added in series so as to raise the overall p.f. to unity |
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Answer» Inductive Impedance V1 I cos φ1 = power; 230 × I1 × 0.5 = 100 ; I1 = 0.87 A Now, I12 R1 = power or 0.872 R1 = 100; R1 = 132 Ω ; Z1 = 230/0.87 = 264 Ω XL = √(Z21 - R21) = √(2642 - 1322) = 229Ω Capacitance Impedance I2 = 60/230 × 0.6 = 0.434 A ; R2 = 60/0.4342 = 318 Ω Z2 = 230/0.434 = 530 Ω ; XC = √(5302 - 3182) = 424Ω (capacitive) When Z1 and Z2 are connected in series R = R1 + R2 = 132 + 318 = 450 Ω; X = 229 − 424 = − 195 Ω (capacitive) Z = √(R2 + X2) = √(4502 + (-195)2) = 490 , 230 / 490 0.47 A (i) Total power absorbed = I2R = 0.472 × 450 = 99 W, cos φ = R/Z = 450/490 = 0.92 (lead) (ii) Power factor will become unity when the net capacitive reactance is neutralised by an equal inductive reactance. The reactance of the required series pure inductive coil is 195 Ω. |
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