Asked by abdul rahman
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?
(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?
Answers
Answered by
Henry
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.
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.
Answered by
abdul rahman
Plz answer b part......(C) n (d) part are wrong !
Answered by
Anonymous
wrong answer
Answered by
Anonymous
please abdul answer of this question.
Answered by
FLu
All of the answers are wrong! Please help!
Answered by
FLu
sorry answer a) seems right the rest appears wrong though. Anyone for b),c),d)?
Answered by
abdul rahman
plz do u hav answers for other edx HW Q's ????
Answered by
FLu
Did you figure the above problem out Abdul?
Thank
Thank
Answered by
FLu
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
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
Answered by
FLu
I meant divide 12/0.09 and then multiply the result by 0.95 which is approx. 126
b) and c) anyone?
b) and c) anyone?
Answered by
FLu
ANyone for Problem 2: Displacement Current a) please?
b) 0
Open Switch on an RL Circuit
b) and c) please?
b) 0
Open Switch on an RL Circuit
b) and c) please?
Answered by
abdul rahman
flu do u hav answers for Q2 part a
Q3 part b and part c, Q5 and Q7 all parts ???
Q3 part b and part c, Q5 and Q7 all parts ???
Answered by
My
Anyone for this please?
Problem 2: Displacement Current a) please?
Problem: Open Switch on an RL Circuit
b) and c) please?
Problem 2: Displacement Current a) please?
Problem: Open Switch on an RL Circuit
b) and c) please?
Answered by
Ira
Yes, please the question below too?!
Problem 2: Displacement Current a)
Problem: Open Switch on an RL Circuit
b) and c)
Problem 2: Displacement Current a)
Problem: Open Switch on an RL Circuit
b) and c)
Answered by
FLu
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.
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.
Answered by
dd
i need only the c1
Answered by
ev
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
C1: B + C2
Answered by
Supraconductor
remember that E=1/2.L.I² for b)
Answered by
Ur
Anybody figure out c)?
Answered by
FLu
Anyone for Problem
Opening a Switch on an RL Circuit
c) please?
Opening a Switch on an RL Circuit
c) please?
Answered by
Supraconductor
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)
now you have d) you have just to do answer c) = b) + d)
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