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A resistor and an inductor are connected in series to a battery with emf 240 V and negligible internal resistance. The circuit is completed at time t=0. At a later time t=T the current is 5.00 A and is increasing at a rate of 20.0 A/s. After a long time the current in the circuit is 20.0 A.
What is the value of the time when the current is 7.00 A?

Sagot :

nuhulawal20

Answer:

The value of time T₁ is 0.323 sec

Explanation:

Given the data in the question;

The inductor will be shorted after a long time;

I = V / R

given that; I = 20.0 A and V = 240 V

we substitute

20 = 240 / R

20R = 240

R = 240 / 20

R = 12 Ohms

Now, when I = 5.00 A

voltage across inductor = 240 - ( 5.0 × 12 )

voltage across inductor = 180 V

so

180 = L × di/dt

given that; increasing at a rate of 20.0 A/s

180 = L × 20

L = 180 / 20

L = 9 H

Now, for the time constant of circuit is T₁

T₁ = L / R = 9 / 12

T₁ = 0.75 s

so

I = l₀( 1 - [tex]e^{(-T/T_1)[/tex] )

we substitute

7.00 = 20.0( 1 - [tex]e^{(-T/0.75)[/tex] )

7/20 = 1 - [tex]e^{(-T/0.75)[/tex]

0.35 = 1 - [tex]e^{(-T/0.75)[/tex]

0.35 - 1 =  - [tex]e^{(-T/0.75)[/tex]

-0.65 = - [tex]e^{(-T/0.75)[/tex]

0.65 = [tex]e^{(-T/0.75)[/tex]

ln(0.65) = -T₁ / 0.75

-0.43078 = -T₁ / 0.75

-T₁ = -0.43078 × 0.75

-T₁ = -0.323

T₁ = 0.323 sec

Therefore, The value of time T₁ is 0.323 sec

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