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Inductive Reactance

In an L–R cir...

Question

In an L–R circuit $L = \frac{0.4 H}{π}$ and R = 30 $Ω$ If the circuit has an alternating emf of 220 volt and its frequency 50 cycles per sec, the impendence and current in the circuit respectively will be

50 Ω, 4.4 A

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40.4 Ω, 5 A

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3.07 Ω, 6.0 A

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11.4 Ω, 17.5 A

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Solution

The correct option is A $50 Ω, 4.4 A$
$Z = \sqrt{R^{2} + [L W]^{2}} = \sqrt{30^{2} + {[\frac{0.4}{π} 3 π f]}^{2}}$
$Z = \sqrt{30^{2} + (0.8 \times 50)^{2}} = \sqrt{30^{2} + 40^{2}} = 50 Ω$
$I = \frac{V}{Z} = \frac{220}{50} = 4.4 A$

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In an L–R circuit L=0.4Hπ and R = 30 Ω If the circuit has an alternating emf of 220 volt and its frequency 50 cycles per sec, the impendence and current in the circuit respectively will be

The correct option is A 50 Ω, 4.4AZ=√R2+[LW]2=√302+[0.4π3πf]2 Z=√302+(0.8×50)2=√302+402=50Ω I=VZ=22050=4.4A

In an L–R circuit $L = \frac{0.4 H}{π}$ and R = 30 $Ω$ If the circuit has an alternating emf of 220 volt and its frequency 50 cycles per sec, the impendence and current in the circuit respectively will be

The correct option is A $50 Ω, 4.4 A$
$Z = \sqrt{R^{2} + [L W]^{2}} = \sqrt{30^{2} + {[\frac{0.4}{π} 3 π f]}^{2}}$
$Z = \sqrt{30^{2} + (0.8 \times 50)^{2}} = \sqrt{30^{2} + 40^{2}} = 50 Ω$
$I = \frac{V}{Z} = \frac{220}{50} = 4.4 A$