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?
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.
Plz answer b part......(C) n (d) part are wrong !
wrong answer
please abdul answer of this question.
All of the answers are wrong! Please help!
sorry answer a) seems right the rest appears wrong though. Anyone for b),c),d)?
plz do u hav answers for other edx HW Q's ????
Did you figure the above problem out Abdul?
Thank
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
I meant divide 12/0.09 and then multiply the result by 0.95 which is approx. 126
b) and c) anyone?