Question

A coil has an inductance of $\frac{2.2}{\pi} \mathrm{H}$ and is joined in series with a resistance of $220 \Omega$. When an alternating emf of $220 \mathrm{~V}$ at $50 \mathrm{~c.p.s}$ is applied to it, then the wattless component of the rms current in the circuit is:

A 0.5 $\mathrm{~A}$ B 0.7 $\mathrm{~A}$ C 5 $\mathrm{~A}$ D 7 $\mathrm{~A}$

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Answer

Question A coil has an inductance of $\frac{2.2}{\pi} \ \mathrm{H}$ and is joined in series with a resistance of $220 \ \Omega$. When an alternating emf of $220 \ \text{V}$ at $50 \ \text{c.p.s}$ is applied to it, then the wattless component of the rms current in the circuit is:

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

The correct answer is A) 0.5 A.

Given:

  • Inductance, $L = \frac{2.2}{\pi} \ \mathrm{H}$
  • Resistance, $R = 220 \ \Omega$
  • Alternating emf, $V = 220 \ \text{V}$
  • Frequency, $f = 50 \ \text{Hz}$

First, we calculate the inductive reactance, $X_L$: $$ X_L = 2 \pi f L = 2 \pi \cdot 50 \cdot \frac{2.2}{\pi} = 220 \ \Omega $$

Next, we find the impedance, $Z$, of the series circuit: $$ Z = \sqrt{R^2 + X_L^2} = \sqrt{220^2 + 220^2} = 220 \sqrt{2} \ \Omega $$

The rms current, $I$, in the circuit can be given by Ohm's law for AC circuits: $$ I = \frac{V}{Z} = \frac{220}{220 \sqrt{2}} = \frac{1}{\sqrt{2}} \ \text{A} $$

To find the wattless (or reactive) component of the current, $I \sin \phi$, we use the ratio of the inductive reactance to the impedance: $$ I \sin \phi = \frac{V}{Z} \cdot \frac{X_L}{Z} = \frac{220}{220 \sqrt{2}} \cdot \frac{220}{220 \sqrt{2}} = \frac{1}{2} = 0.5 \ \text{A} $$

Thus, the wattless component of the rms current in the circuit is 0.5 A.


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