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What is the resistance of the inductive coil that takes 5A current across a 240V, 50Hz supply at a 0.8 power factor?

A) 48
B) 42.5
C) 38.4
D) 26.6

Answer :

To determine the resistance of the inductive coil, we must first understand how the properties of the coil, voltage, current, and power factor relate to resistance.

The power factor (cos [tex]\phi[/tex]) tells us how effectively the electrical power is being converted into useful work. A power factor of 0.8 means that 80% of the power is being used effectively.

The apparent power ([tex]S[/tex]) is given by the product of the voltage ([tex]V[/tex]) and the current ([tex]I[/tex]). So:

[tex]S = V \times I = 240 \times 5 = 1200 \text{ VA}[/tex]

The real power ([tex]P[/tex]) consumed by the coil is the product of the apparent power and the power factor:

[tex]P = S \times \, \text{power factor} = 1200 \times 0.8 = 960 \text{ W}[/tex]

Since the real power [tex]P[/tex] can also be expressed in terms of resistance [tex]R[/tex] and current [tex]I[/tex] as [tex]P = I^2 \times R[/tex], we can solve for [tex]R[/tex]:

[tex]960 = 5^2 \times R[/tex]
[tex]960 = 25R[/tex]
[tex]R = \frac{960}{25} = 38.4 \, \Omega[/tex]

Therefore, the resistance of the inductive coil is [tex]38.4 \, \Omega[/tex].

The correct multiple-choice option is C) 38.4.

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