Question

a coil taks a current of 0.6 lagging factor from a 220v 60 hz songle phase source .if the coil is modeled by a eeries rl circuit.find the complex power in the coil and the value of r and l​

Answer 1

Answer:

If the two circuits are connected in series across a 230 V 50 Hz ... A circuit takes a current of 3A at a power factor of 0.6 lagging ... (c) Power factor =cosϕ=RZ=39.2641.92= 0.939

Answer 2

Complete question:

A coil takes a current of 1 A at 0.6 lagging power factor from a 220 V, 60 Hz single phase source, If the coil is modeled by a series RL circuit, find, (i) The complex power in the coil and (ii) The values of R and L.

Answer:

(i) The complex power is 132 + 172j.

(ii) The value of R and L is 132 and 0.466 H

Explanation:

Given,

The current flowing in the coil (i) is 1 A.

The lagging power factor cos(φ) = 0.6

The voltage supplied is (V) = 220 V.

The frequency maintained in the coil is f = 60 Hz.

To find,

(i) The complex power in the coil.

(ii) The values of R and L.

Calculation,

Let the resistance of the coil be 'R', and the inductance of the coil be 'L'.

Let the resistance of the entire coil be 'Z'.

From ohms law:

Z = V/i

⇒ Z = 220/1

⇒ Z = 220 V.

As cos(φ) = R/Z

⇒ 0.6 = R/220

⇒ R = 132 ohms.

As the coil is having a lagging power factor, the circuit is having an inductor.

Hence,

$X_L = \sqrt{Z^2 - R^2}$

$\implies X_L = \sqrt{220^2 - 132^2}$

$X_L$ = 176 ohms.

Inductive reactance ($X_L$ ) = 2πfL

176 = 2 × π × 60 × L

L = 0.466 H

(i) Complex power (S) = Real power(P) + j × Apparent power(Q)

S = P + jQ

⇒ S = VI cos(φ) + j(VIsin(φ))

⇒ S = 132 + j176

Therefore, the complex power is 132 + 172j

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