Question

The potential difference measured across a coil is 4.5v, when it carries a direct current of 9A. The same coil when carries an alternating current of 9A at 25 Hz, the potential difference is 24v. Find the current, the power factor and the power when supplied by 50v, 50 Hz supply.

Answer 1

Answer:

Power factor of the coil is 0.095

current in the coil is 9.5 A

Power of the coil is 45.15 W

Explanation:

When Direct current flows through the coil then we have

$V = iR$

so we have

$4.5 = 9 R$

$R = 0.5 ohm$

now when AC current flows through the coil then we have

$V = i z$

$24 = 9 z$

$z = 2.67 ohm$

now we know that

$z = \sqrt{x_L^2 + R^2}$

$2.67^2 = x_L^2 + 0.5^2$

$x_L = 2.62 ohm$

We know that

$x_L = 2\pi f L$

$2.62 = 2\pi (25) L$

$L = 0.0167 H$

now we use 50 Hz 50 V AC supply so we have

$x_L = (2\pi)(50)(0.0167)$

$x_L = 5.24 ohm$

now impedence of the circuit is given as

$z = \sqrt{x_L^2 + R^2}$

$z = 5.26 ohm$

i) Power factor of the coil is given as

$cos\phi = \frac{R}{z}$

$cos\phi = \frac{0.5}{5.26}$

$cos\phi = 0.095$

ii) Current in the circuit is given as

$i = \frac{50}{z}$

$i = \frac{50}{5.26} = 9.5 A$

iii) Power used by the coil is given as

$P = i v cos\phi$

$P = (9.5)(50)(0.095)$

$P = 45.15 W$

#Learn

Topic : AC circuit

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