Question

An inductor of 2 h and a resistance of 10 ohm are connected to a battery of 5 v in series. The initial rate of change of current i

Answer 1

Solution is given in the picture.

Thus, The initial rate of change of current 2.5 m/s.

Answer 2

Answer:

The initial rate of change of current  is 2.5 A/s.

Explanation:

Given that,

Inductor = 2 h

Resistance = 10 ohm

Voltage = 5 V

We need to calculate the initial rate of change of current i

Using formula of current

$i=i_{0}-e^{\dfrac{-Rt}{L}}$...(I)

On differentiating equation (I)

$\dfrac{di}{dt}=\dfrac{d}{dt}i_{0}e^{\dfrac{-Rt}{L}}$

$\dfrac{di}{dt}=0+\dfrac{i_{0}R}{L}e^{\dfrac{Rt}{L}}$

Initially , t = 0

$\dfrac{di}{dt}=\dfrac{i_{0}R}{L}$

$\dfrac{di}{dt}=\dfrac{V}{L}$

Put the value into the formula

$\dfrac{di}{dt}=\dfrac{5}{2}$

$\dfrac{di}{dt}=2.5\ A/s$

Hence, The initial rate of change of current  is 2.5 A/s.