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

Answer:

1.2 A

Explanation:

We are given that

Time, dt=78 ms=$78\times 10^{-3}s$

$1 ms=10^{-3} s$

$I_s=4.1mA=4.1\times 10^{-3} A$

$1 mA=10^{-3}A$

$R=12\Omega$

$M=3.2mH=3.2\times 10^{-3} H$

We have to find the change in the primary current.

$V_s=I_sR=4.1\times 10^{-3}\times 12=49.2\times 10^{-3} V$

$V_s=M\frac{dI}{dt}$

$dI=\frac{V_sdt}{M}=\frac{49.2\times 10^{-3}\times 78\times 10^{-3}}{3.2\times 10^{-3}}$

$dI=1.2 A$