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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 3.00 A and is increasing at a rate of 20.0 A/s. After a long time the current in the circuit is 25.0 A.

Respuesta :

Answer:

T = 0.141 s

Explanation:

given,

Battery emf = 240 V

Current after long time = 25 A

increasing rate of current = 20 A/s

time = ?

after long time

V = I R

[tex]R = \dfrac{V}{I}[/tex]

[tex]R = \dfrac{240}{25}[/tex]

R = 9.6 Ω

Voltage across inductor when I = 3 A

Voltage = 240 - 3 x 9.6

Voltage across inductor = 211.2 V

[tex]V = L \dfrac{di}{dt}[/tex]

[tex]211.2 = L \times 20[/tex]

L = 10.56 H

time constant

[tex]\tau = \dfrac{L}{R}[/tex]

[tex]\tau = \dfrac{10.56}{9.6}[/tex]

[tex]\tau =1.1\ s[/tex]

now,

[tex]i = i_0 (1 - e^{\dfrac{-T}{\tau}})[/tex]

[tex]3 = 25(1 - e^{\dfrac{-T}{1.1}})[/tex]

[tex]e^{\dfrac{-T}{1.1}}= 0.88[/tex]

taking log both side

[tex]\dfrac{-T}{1.1}= -0.128[/tex]

   T = 0.141 s

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