30.19 An inductor with an inductance of 2.80 H and a resistance of 7.30 is conne

Question

30.19

An inductor with an inductance of 2.80 H and a resistance of 7.30 is connected to the terminals of a battery with an emf of 6.50 V and negligible internal resistance.

Part A

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

2.32

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Correct

Part B

Find the rate of increase of current at the instant when the current is 0.500 A .

.78

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Part C

Find the current 0.270 s after the circuit is closed.

.42

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Part D

Find the final steady-state current.

30.19

An inductor with an inductance of 2.80 H and a resistance of 7.30 is connected to the terminals of a battery with an emf of 6.50 V and negligible internal resistance.

Part A

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

didt =

2.32

A/s

SubmitMy AnswersGive Up

Correct

Part B

Find the rate of increase of current at the instant when the current is 0.500 A .

didt=

.78

A/s

SubmitMy AnswersGive Up

Incorrect; Try Again; 5 attempts remaining

Part C

Find the current 0.270 s after the circuit is closed.

I=

.42

A

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Incorrect; Try Again; 5 attempts remaining

Part D

Find the final steady-state current.

I= A

Explanation / Answer

B)

i = current in the circuit = 0.5 A

Vr = Voltage across the resistor

R = resistance = 7.30 ohm

using ohm's law

Vr = iR

Vr = (0.5) (7.30)

Vr = 3.65 Volts

VL = Voltage across inductor = V - Vr = 6.50 - 3.65 = 2.85 volts

L = inductance = 2.80 H

rate of increase of current is given as

di/dt = VL /L = 2.85 /2.80 = 1.02 A/s

c)

T = time constant = L/R = 2.80/7.30 = 0.384 sec

t = time given = 0.270 s

io = maximum current = V/R = 6.50/7.30 = 0.89 A

using the equation

i = io (1 - e-t/T)

i = 0.89 (1 - e-0.270/0.384)

i = 0.45 A

d)

steady state current is given as

io = maximum current = V/R = 6.50/7.30 = 0.89 A

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