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A coil has an inductance of 0.7 H and is connected in series with a resistance of 220 ohms. When an alternating e.m.f. of 220 V at 50 Hz is applied to it, what is the wattless component of the current in the circuit?

A. 5 A
B. 0.5 A
C. 0.7 A
D. 7 A

Answer :

Final answer:

The wattless component of current in an RL circuit can be calculated using the inductive reactance and Ohm's law. The inductive reactance is proportional to frequency and inductance. So, using the provided values, the current, calculated from the formulas, equals 0.7A.

Explanation:

The wattless component of the current in a circuit refers to the current related to the reactive impedance, in this case the conductance due by the inductor. The inductance of the coil and the frequency of the alternating current are the variables required to calculate this so-called 'wattless' or 'reactive' current.

The impedance provided by an inductor to an AC current is called inductive reactance (XL), and it is calculated by the formula:

XL = 2πfL

with:

  • f being the frequency of the AC voltage source in hertz (in your case, 50 hertz or cycles per second), and
  • L being the inductance (in this case, 0.7 H).

After calculating the inductive reactance, the component of current that is in quadrature opposition (or 90 degree phase shift) to the voltage can be calculated using Ohm's law, by dividing the voltage by the resistor and the inductive reactance.

So, the wattless component of the current, I, as you asked, would be determined by the following equation:

I = V / sqrt((R * R) + (XL * XL))

Considering the given values from your question, XL = 2nfL = 2*3.14*50*0.7 = 220 ohms. Then, I = 220 / sqrt((220 * 220) + (220 * 220)) = 0.7 Amps. So, the right answer would be 0.7A.

Learn more about RL Circuits here:

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