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

Question

Katelynn Kane

Engineering

9 September, 15:33

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

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Answers (1)

Know the Answer?

Jamiya Madden · Answer 1

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