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If number of turns in primary and secondary coils is increased to two times each, the mutual inductane .....

Answer»

becomes FOUR TIMES
becomes two times
becomes `1/4` times
remains unchanged

Solution :`N._1=2N_1,N._2=2N_2`
MUTUAL INDUCTANCE =`M=phi/I=(mu_0 N_1N_2A)/L`
`therefore M prop N_1N_2`
`therefore M_2/M_1=(N._1N._2)/(N_1N_2)=((2N_1)(2N_2))/(N_1N_2)=4`
`therefore M_2=4M_1`
`therefore` Mutual inductance becomes four times.


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