Two inductors 0.4 H and 20.6 H are connected in parallel. If this combination is connected in series with an inductor of inductance 0.76 H. The equivalent inductance of the circuit will be (nearly)
A
2 H
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B
1 H
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C
0.2 H
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D
0.1 H
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Solution
The correct option is B 1 H L1=L1L2L1+L2+L3 =0.4×20.60.4+20.6+0.76 =8.2421+0.76 =0.39+0.76 =1.15 ≈1H