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) ?
How much energy (in Joules) has been dissipated in the resistor?
4 answers
b. I = E/R = 12/0.05 = 240A, =
Please ignore my 1st response. My
computer malfunctioned. I will have to
repeat the entire process.
computer malfunctioned. I will have to
repeat the entire process.
a. I = E/R = 12/0.05 = 240A.
0.95I = 0.95*240 = 228A.
Vr + Vc = 12
228*0.05 + 12/e^(t/T) = 12
11.4 + 12/e^(t/T) = 12
12/e^(t/T) = 12-11.4 = 0.6
e^(t/T) = 12/0.6 = 20
t/T = 3.
T = L/R = 0.09/0.05 = 1.8=Time constant.
t/1.8 = 3
t = 5.4 s.
b. Energy = 0.5L*I^2=.5*0.09*(228)^2 =
2339.3 J.
0.95I = 0.95*240 = 228A.
Vr + Vc = 12
228*0.05 + 12/e^(t/T) = 12
11.4 + 12/e^(t/T) = 12
12/e^(t/T) = 12-11.4 = 0.6
e^(t/T) = 12/0.6 = 20
t/T = 3.
T = L/R = 0.09/0.05 = 1.8=Time constant.
t/1.8 = 3
t = 5.4 s.
b. Energy = 0.5L*I^2=.5*0.09*(228)^2 =
2339.3 J.
thanks, do you have the c?
1 answer