• # question_answer Two coils A and B having turns 300 and 600 respectively are placed near each other. On passing a current of 3.0 ampere in A, the flux linked with A is $1.2\times {{10}^{-4}}$ Wb and with B it is $9.0\times {{10}^{-5}}$Wb. The mutual inductance of the system is: A) $4\times {{10}^{-5}}H$                 B) $3\times {{10}^{-5}}H$     C)   $2\times {{10}^{-5}}H$     D)   $1.8\times {{10}^{-2}}H$

Correct Answer: D

Solution :

$\left| {{e}_{s}} \right|={{N}_{s}}\frac{d{{\Phi }_{s}}}{dt}$and $\left| {{e}_{s}} \right|=M\frac{d{{I}_{s}}}{dt}\,\therefore \,{{N}_{s}}\frac{d{{\Phi }_{s}}}{dt}=M\frac{d{{I}_{s}}}{dt}$ $M={{N}_{s}}\frac{d\Phi }{d{{I}_{p}}}=\frac{600\times 9.00\times {{10}^{-5}}}{3.0}=1800\times {{10}^{-5}}H.$ Hence, the correction option is(d).

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