# physics

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Consider a circuit with an RL series with L=0.09 H and R=0.05 Ohm. At t=0 the circuit is connected to a battery which provides V0=12 V.

(a)How long does it take to the current to equal a fraction 0.95 of the steady state current?

(b) What is the energy stored (in Joules) in the magnetic field when the current equals a fraction 0.95 of the steady state current?

(c) What is the total energy delivered (in Joules) by the battery up to the time t found in part (a) ?

(d) How much energy (in Joules) has been dissipated in the resistor?

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I = E/R = 12/0.05 = 24q Amps = Steady-state current.

0.95I = 0.95 * 240 = 228A.

Vr + Vc = 12 Volts
228*00.05 + Vc = 12
Vc = 12-11.4 = 0.6 Volts = Voltage across coil when i = 0.95I steady-state

a. Vc = 12/e^(t/T) = 0.6
e^(t/T) = 12/0.6 = 20
t/T = Ln20 = 3.0
T = L/R = 0.09/0.05 = 1.8 s.
t/1.8 = 3
t = 5.4 s.

c. Energy = E*I*t = 12*228*5.4=14,774.4
Joules.

d. Energy = I^2*R*t = (228)^2*0.05*5.4 =
14,036 Joules.

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Plz answer b part......(C) n (d) part are wrong !

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sorry answer a) seems right the rest appears wrong though. Anyone for b),c),d)?

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plz do u hav answers for other edx HW Q's ????

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Did you figure the above problem out Abdul?
Thank

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I got d) 8165
you multiply 12/0.09
and than apply what Henry provided for answer c) 12*126*5.4=8165
It seems the numbers are getting mixed up, can someone figure out b) and c) with it?
thanks

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I meant divide 12/0.09 and then multiply the result by 0.95 which is approx. 126

b) and c) anyone?

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ANyone for Problem 2: Displacement Current a) please?
b) 0

Open Switch on an RL Circuit

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flu do u hav answers for Q2 part a
Q3 part b and part c, Q5 and Q7 all parts ???

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Problem 2: Displacement Current a) please?

Problem: Open Switch on an RL Circuit

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Yes, please the question below too?!

Problem 2: Displacement Current a)

Problem: Open Switch on an RL Circuit
b) and c)

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Can somebody help with Problem: Displacement Current please?

I thought the formula is:
(1.25663706*10^-6*0.04*0.2*0.0705)/(2*pi*0.05^2)

However, I am not getting the correctr answer.

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i need only the c1

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C2: integral 1 to 5.4 of (12/0.05)^2)* (1-e^(-(0.05 t)/0.09))^2 * 0.05 dt

C1: B + C2

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remember that E=1/2.L.I² for b)

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Anybody figure out c)?

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Anyone for Problem
Opening a Switch on an RL Circuit

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Flu > ev give you the answer d (just integrate what he wrote)

now you have d) you have just to do answer c) = b) + d)

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