Question

Two impedances Z1 = (8+j10 )Ω and Z2 = (7+j9)Ω are connected in parallel. An impedance of Z3 = (5-j2)Ω is connected in series with the parallel combination of Z1 and Z2. A voltage of 200V is applied across the entire combination. Determine the current and power drawn by the circuit

Answer 1

Explanation:

Illustrate with the help of any two examples that

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women, has been duly taken care of in local bodies

Answer 2

Answer:

Let I = 15 ∠ 0º ; Z1 = 10 + j15 = 18 ∠ 57º

Z2 = 6 −j8 = 10 ∠ − 53.1º

Total impedance, Z = (Z1Z2)/(Z1 + Z2) = ((10 + j15)(6 - j8))/(16 + j7)

Applied voltage is given by V = IZ = 15 ∠ 0º × 10.3 ∠−20.4º = 154.4 ∠− 20.4º

I1 = V/Z1 = 154.5 ∠ − 20.4º/18 ∠ 57º = 8.58 ∠ − 77.4º

I2 = V/Z2 = 154.5 ∠−20.4º/10 ∠−53.1º = 15.45 ∠32.7º

We could also find branch currents as under :

I1 = I. Z2/(Z1 + Z2) and I2 = I.Z1/(Z1 + Z2)

I1 lags behind V by (77.4º − 20.4º) = 57º and

I2 leads it by (32.7º + 20.4º) = 53.1º. ∴ P1 = I12R1 = 8.582 × 10 = 736 W; p.f. = cos 57º = 0.544 (lag) P2 = I22 R2 = 15.452 × 6 = 1432 W ; p.f. = cos 53.1º = 0.6 Combined p.f. = cos 20.4º = 0.937 (lead)