Question

During a 78-ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a 4.1-mA current in a nearby secondary coil. The secondary coil is part of a circuit in which the resistance is 12 Ω. The mutual inductance between the two coils is 3.2 mH. What is the change in the primary current?

Answer 1

1.2 A

Step-by-step explanation:

We are given that

Time, dt=78 ms=
`78* 10^(-3)s`

`1 ms=10^(-3) s`

`I_s=4.1mA=4.1* 10^(-3) A`

`1 mA=10^(-3)A`

`R=12\Omega`

`M=3.2mH=3.2* 10^(-3) H`

We have to find the change in the primary current.

`V_s=I_sR=4.1* 10^(-3)* 12=49.2* 10^(-3) V`

`V_s=M(dI)/(dt)`

`dI=(V_sdt)/(M)=(49.2* 10^(-3)* 78* 10^(-3))/(3.2* 10^(-3))`

`dI=1.2 A`