WAEC - Physics (2002 - No. 43)
Calculate the inductance L, of the coil in the circuit diagram shown above
3.8H
0.6H
0.4H
0.2H
Explanation
XL = \(\frac{V}{I} = \frac{240}{2} = 120\Omega\)
L = \(\frac{X_L}{2 \pi f}\)
= \(\frac{120 \times \pi}{2 \times \pi \times 100}\)
= 0.6H
L = \(\frac{X_L}{2 \pi f}\)
= \(\frac{120 \times \pi}{2 \times \pi \times 100}\)
= 0.6H
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