First slide

Phasors and Phasor Diagrams

Question

An ideal choke takes a current of 10 A when connected to an AC supply of 125 volt and 50 Hz. A pure resistor under the same conditions takes a current of 12.5 A. If the two are connected to an AC supply of 100 volt and 40 Hz, then the current in series combination of above resistor and inductor is

Moderate

A

10 A
B

12.5 A
C

20 A
D

25 A

Solution

$R = \frac{125}{12 .5} = 10 Ω$
$X_{L} = ωL = 2 πvL = \frac{V}{I} = \frac{125}{10} = 12 .5$
$∴ 2 {πv}_{1} L = 12 .5$
or $2 πL = \frac{12 .5}{50} = 0 .25$
$∴ X_{L} = 2 πL \times v_{2} = 0 .25 \times 40 = 10 Ω$
Impedance of the circuit
$Z = \sqrt{R^{2} + X_{L}^{2}} = 10 \sqrt{2} Ω$
$∴ Current = \frac{100 \sqrt{2}}{10 \sqrt{2}} = 10 A$

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