Problem 7: An RL circuit is shown on the right. L .SHR 2.5 Otheexpertta Part (a)

Question

Problem 7: An RL circuit is shown on the right. L .SHR 2.5 Otheexpertta Part (a) Switch A is closed att 0. Express the current l in the circuit as a function time in terms of L, R, and E. Schematic Choice tL tR (1 -e RT) I(t) I(t) e I(t) (1-e R Part (b) What's the direction of the current I, counterclockwise or clockwise? Multiple Choice 1) Clockwise 2) Counterclockwise. Part (c) Calculate the numerical value ofi att 0.1 s in amperes. Numeric A numeric value is expected and not an expression. Part (d) Calculate the numerical value of I at t LR s in amperes. Numeric A numeric value is expected and not an expression. Part (e) Calculate the numerical value ofI, in amperes, when t goes to infinity. Numeric Anumeric value is expected and not an expression.

Explanation / Answer

Here ,

L = 4.8 H

R = 144 Ohm

E = 2.5 V

part a)

as switch is closed

the current in inductor is given as

I = E/R(1 - e^(-tR/L))

the correct option is E/R * (1 - e^(-tR/L))

part b)

the direction of current is counter clockwise

part c)

at t = 0.1 s

current , I = 2.5/144 * (1 - e^(-0.1 * 144/(4.8)))

I = 0.0165 A

part d)

at t = L/R

I = E/R(1 - e^(-(L/R) * R/L))

I = 2.5/144 * (1 - e^(-1))

I = 0.011 A

the current I is 0.011 A

part e) at t = infinity

I = E/R

I = 2.5/144 A

I = 0.0174 A

the currnet is 0.0174 A

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