Question

The flux in a closed circuit of resistance 10ohm varies with time according to the equation flux=3t2-t+1,where flux is in Weber and t is in s , value of induced current at t=1 s is?

Answer 1

This will help you mate

Answer 2

The induced current in the circuit is 0.5 A.

Explanation:

Resistance, R = 10 ohms

The flux in a closed circuit of resistance 10 ohm varies with time according to the equation as:

$\phi=3t^2-t+1$

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(3t^2-t+1)}{dt}$

$\epsilon=6t-1$

At t = 1 s

$\epsilon=6(1)-1=5\ V$

Let I is the induced current in the circuit. It is given by :

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{5}{10}$

I = 0.5 A

So, the induced current in the circuit is 0.5 A. Hence, this is the required solution.

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