Question on: WAEC Physics - 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 Option: A
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
= = 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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