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.
