First slide

Inductance

Question

A coil of wire of a certain radius has 600 turns and a self inductance of 108 mH. The self inductance of a 2^nd similar coil of 500 turns will be

Moderate

A

74 mH
B

75 mH
C

76 mH
D

77 mH

Solution

$\frac{L_{B}}{L_{A}} = {(\frac{n_{B}}{n_{A}})}^{2} \Rightarrow L_{B} = {(\frac{500}{600})}^{2} \times 108 = 75 mH$

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