The coefficient of mutual inductance of two circuits A and B is 3mH and their respective resistance is 10Ω and 4Ω. How much current should change in 0.02 s in circuit A, so that the induced current in B should be 0.0060A?
A
0.24A
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B
1.6A
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C
0.18A
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D
0.16A
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Solution
The correct option is B0.16A induced current in B=0.006A=6×10−3A induced emf in B=6×10−3×4V=24×10−3V Now, MdIdt=24×10−3 or, dI=24×10−3×0.023×10−3A=0.16A