Answer:
1.2 A
Explanation:
We are given that
Time, dt=78 ms=[tex]78\times 10^{-3}s[/tex]
[tex]1 ms=10^{-3} s[/tex]
[tex]I_s=4.1mA=4.1\times 10^{-3} A[/tex]
[tex]1 mA=10^{-3}A[/tex]
[tex]R=12\Omega[/tex]
[tex]M=3.2mH=3.2\times 10^{-3} H[/tex]
We have to find the change in the primary current.
[tex]V_s=I_sR=4.1\times 10^{-3}\times 12=49.2\times 10^{-3} V[/tex]
[tex]V_s=M\frac{dI}{dt}[/tex]
[tex]dI=\frac{V_sdt}{M}=\frac{49.2\times 10^{-3}\times 78\times 10^{-3}}{3.2\times 10^{-3}}[/tex]
[tex]dI=1.2 A[/tex]
