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A coil of inductance 0.12 H and resistance 4\(\Omega\), is connected across a 240V, 50Hz...

Physics
WAEC 2013

A coil of inductance 0.12 H and resistance 4\(\Omega\), is connected across a 240V, 50Hz supply. Calculate the current through it. [\(\pi\) = 3.142]
  • A. 6.3A
  • B. 33.3A
  • C. 37.2A
  • D. 40.0A
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Correct Answer: Option A
Explanation
XL = \(2\pi f L\)

= = 2 x 3.142 x 50 x 0.12

= 37.68\(\Omega\)

Z = \(\sqrt{R^2 + X_L^2} = \sqrt{4^2 + 37.68^2}\)

I = \(\frac{V}{z} = \frac{240}{37.89}\)

= 6.3A

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