An RL circuit is shown on the right. L = 2.6 H.R = 315 Ohm. epsilon = 8.5 V. (a)

Question

An RL circuit is shown on the right. L = 2.6 H.R = 315 Ohm. epsilon = 8.5 V. (a) Switch A is closed at t = 0. Express the current I in the circuit as a function of time in terms of L, R, and epsilon. (b) What's the direction of the current I, counterclockwise or clockwise? 0% deduction per feedback. Calculate the numerical value of I at t = 0.1 s in amperes. Calculate the numerical value of I at t = L/R s in amperes. Calculate the numerical value of I, in amperes, when t goes to infinity. After a long time, when the current reaches its steady value, open switch A and close switch B at the same time. Count moment as t = 0. Express the current in the circuit, I, as a function of time in terms of L, R, and e. (g) Now what is the direction of the current I, counter-clockwise or clockwise? (h) Calculate the numerical value of I at t = 0.1 s in amperes. (i) Calculate the numerical value of I at t = L/R s in amperes. (j) Calculate current, in amperes, after an infinite time.

Explanation / Answer

a) we know that

during charging of LR circuit

I(t) = E/R [ 1 - e-Rt/L ]

b) direction will be anti CW

c) when t = 0.1 s then I = ?

we know that

I(0.1) = E/R [ 1 - e-Rt/L ]

= 8.5/315 [ 1 - e-315x0.1/2.6 ]

= 0.027 A

d) t = L/R

so

I(t) = E/R [ 1 - e-Rt/L ]

t = L/R

I(t) = 8.5/315 [ 1 - e-1 ]  

= 0.017 A

e) t--------> infinite

then

I(t) = E/R [ 1 - e-Rt/L ]

I(t) = E/R [ 1 - 0]

I(t) = E/R

I(t) = 8.5/315

= 0.0269 A

f) now after opening A and closing B then circuit will be work as discharging of inductor

now the current formula will be

I(t) = E/R x e-Rt/L  

g) during discharging the current will flow clockwise

h) at t = 0.1 s

I = ?

I(t) = E/R x e-Rt/L  

I(t) = 8.5/315 x e-315x0.1/2.6

I(t) = 1.48 x 10-7 A  

i) when t = L/R

I(t) = E/R x e-RL/LR  

I(t) = 8.5/315 x e-1

I(t) = 9.92 x10-3 A

j) after t -------> infinite

I(t) = E/R x e-Rt/L  

I(t) = 0

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