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

(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

Put the value into the formula

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

....(I)

Where,

Using formula of current

The rate of current increases is

Put the value in the equation (I)

The rate of current increases is 0.8658 A/s.

(c). We need to calculate the current

Using formula of current

Put the value into the formula

The current is 0.435 A.

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

Using formula of steady state

The final steady-state current is 0.75 A.

Hence, This is the required solution.

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