Question

An inductor with an inductance of 2.30H and a resistance of 8.00 Ω isconnected to the terminals of a battery with an emf of 6.00 V andnegligible internal resistance.
(a) Find the initial rate of increase ofcurrent in the circuit.
1 A/s

(b) Find the rate of increase of current at the instant when thecurrent is 0.500 A.
2 A/s

(c) Find the current 0.250 s after the circuit is closed.
3 A

(d) Find the final steady-state current.
4 A

Answer 1

Answer:

(a). The initial rate is 2.60 A/s.

(b). The rate of current increases is 0.8658 A/s.

(c). The current is 0.435 A.

(d). The final steady-state current is 0.75 A.

Explanation:

Given that,

Inductance = 2.30 H

Resistance = 8.00 Ω

Voltage = 6.00 V

(a). We need to calculate the initial rate of increase of current in the circuit.

Using formula of initial rate

$V=initial\ rate\times inductance$

$initial\ rate=\dfrac{V}{L}$

Put the value into the formula

$initial \rate=\dfrac{6.00}{2.30}$

$initial\ rate=2.60\ A/s$

The initial rate is 2.60 A/s.

(b). We need to calculate the rate of increase of current at the instant when the current is 0.500

Using formula of rate of increase of current

$rate\ of \ current\ increase=initial\ rate\times e^{\dfrac{-t}{T}}$....(I)

Where, $T=\dfrac{L}{R}$

$T=\dfrac{2.30}{8.00}$

$T=0.2875$

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

$e^{\dfrac{-t}{T}}=1-(i\times\dfrac{R}{V})$

$e^{\dfrac{-t}{T}}=1-(0.500\times\dfrac{8.00}{6.00})$

$e^{\dfrac{-t}{T}}=0.333$

The rate of current increases is

Put the value in the equation (I)

$rate\ of\ current\ increase=2.60\times0.333$

$rate\ of\ current\ increase=0.8658\ A/s$

The rate of current increases is 0.8658 A/s.

(c). We need to calculate the current

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

Put the value into the formula

$i=\dfrac{6.00}{8.00}\times(1-e^{\dfrac{-0.250}{0.2875}})$

$i=0.435\ A$

The current is 0.435 A.

(d). We need to calculate the final steady-state current

Using formula of steady state

$i=\dfrac{V}{R}$

$i=\dfrac{6.00}{8.00}$

$i=0.75\ A$

The final steady-state current is 0.75 A.

Hence, This is the required solution.