A stationary conductive loop with an internal resistance of 0.5 Ω is placed in a time varying magnetic field. When the loop is c

Question

3 years ago

11

A stationary conductive loop with an internal resistance of 0.5 Ω is placed in a time varying magnetic field. When the loop is c

losed, a current of 5 A flows through it. What will the current be if the loop is opened to create a small gap and a 2-Ω resistor is connected across its open ends?

Engineering

1 answer:

Setler [38] · Answer 1 · 2022-04-21T00:25:37+03:00

Setler [38]3 years ago

5 0

Answer:

e= -AdB/dt

5*0.5 = -AdB/dt

i(2+0.5) = -AdB/dt

from above expressions

2.5 = i2.5

so i = 1A

More Answers:

v=0.5x5= 2.5

2.5/(2+0.5) = 1 A