An inductor with an inductance of 2.10 H and a resistance of 7.60 is connected t

Question

An inductor with an inductance of 2.10 H and a resistance of 7.60 is connected to the terminals of a battery with an emf of 6.30 V and negligible internal resistance.

Part A

Find the initial rate of increase of current in the circuit.

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Part B

Find the rate of increase of current at the instant when the current is 0.550 A .

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Part C

Find the current 0.290 s after the circuit is closed.

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Part D

Find the final steady-state current.

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An inductor with an inductance of 2.10 H and a resistance of 7.60 is connected to the terminals of a battery with an emf of 6.30 V and negligible internal resistance.

Part A

Find the initial rate of increase of current in the circuit.

didt = A/s

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Part B

Find the rate of increase of current at the instant when the current is 0.550 A .

didt = A/s

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Part C

Find the current 0.290 s after the circuit is closed.

I = A

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Part D

Find the final steady-state current.

I = A

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Explanation / Answer

ANSWER

* Given:

L = 2.10 H

R = 7.60 ohms

V = 6.30 v

Part a)

V = L * di/dt
6.30 V = 2.10 * di/dt
di/dt = 3 A/s   ANS.

Part b)

V across resistor when current is 0.550 A: 0.550 * 7.6 = 4.18 V

V across inductor = 6.3 - 4.18 = 2.12 V

2.12 = L * di/dt
di/dt = 1.01 A/s   ANS.

Part c)

You can treat this as a series RL problem. Thus, the differential equation is:
L di/dt + iR = E
di/dt = 1/L (E - iR)
di/ (E - iR) = 1/L dt
integrating both sides, you get,
(E - i(t)*R)/ (E - i(0) * R) = Rt/L
taking natural logs of both sides, and rearranging
E - i(t) * R = (E - i(0)*R) e^(-Rt/L)
i(0) = 0
rearranging,
i(t) = E/R(1-e^-(Rt/L))

i(0.290 s) = 6.3/7.6 * (1- e^(-0.290 * 7.6/2.1)) = 0.54 A ANS.

Part d)

Finally, steady state current, inductor is a short, so its just V/R = 6.3/7.6 = 0.829 A (as t-> infinity, the above equation just becomes E/R).

Regards!!!

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