1.

A coil with resistance 8 Omegaand having 8 turns is connected with a galvanometer having resistance 8 times the resistance of coil. In 4 ms, if magnetic flux linked with this loop changes from 12xx10^(-5) Wb to 18xx10^(-5)Wb then current induced in the loop will be ......

Answer»

1.6 A
`1.6xx10^(-6)` A
`1.6xx10^(-3)` A
`1.6xx10^(-4)` A

Solution :Here `Deltaphi=phi_2-phi_1=18xx10^(-5)-12xx10^(-5)`
`=6xx10^(-5)` Wb
N=8
`R=8 Omega, G=8xx8=64 Omega`
`t=4xx10^(-3)` s
`THEREFORE` Induced current,
`I=epsilon/R` ,
`=(N Delta PHI)/(Deltat)xx1/(R+G)` [Where R.=R+G]
`=(8xx6xx10^(-5))/(4xx10^(-3)xx72)`
`=0.166xx10^(-2)`
`therefore I approx 1.6xx10^(-3)` A


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