In the circuit we have E=48V, R=6 ohm and L=4H. We close the switch at the instant t=0, the current I in the circuit grows from zero to its maximum value Io, and the permanent phase is established.
A) determine the value of Io.
B) calculate the values of the potential difference across R and across the coil at the instant t where I=5A.
C) calculate the potential difference across the coil for i=8A. Explain this result
2 answers
I just need to know what formula i need to use
I assume in series
E = i R + L di/dt
let i = a (1- e^-kt)
then di/dt = a*k e^-kt
when t =oo , E=i R
i = E/R = a
so a = E/R
so
i = (E/R)(1-e^-kt)
di/dt = (E/R) k e^-kt
when t-->0, i = 0 and di/dt = a k
so
E = L di/dt = L a k = L(E/R)k
so
k = R/L
so
In the end all that we need is:
i = (E/R)(1 - e^-Rt/L)
di/dt = (E/R)(R/L)e^-Rt/L
A))Io is current when t-->oo
= E/R (1) = 48/6 = 8 amps
B) i = 8 (1-e^-1.5 t)
Vresistor = 6 i
you can put t = 5 in
Vcoil = 48 - Vresistor but check with
V = L di/dt :)
C) LOL, no di/dt at t=oo so no V across coil
The entire 48 volts is across resistor b then
E = i R + L di/dt
let i = a (1- e^-kt)
then di/dt = a*k e^-kt
when t =oo , E=i R
i = E/R = a
so a = E/R
so
i = (E/R)(1-e^-kt)
di/dt = (E/R) k e^-kt
when t-->0, i = 0 and di/dt = a k
so
E = L di/dt = L a k = L(E/R)k
so
k = R/L
so
In the end all that we need is:
i = (E/R)(1 - e^-Rt/L)
di/dt = (E/R)(R/L)e^-Rt/L
A))Io is current when t-->oo
= E/R (1) = 48/6 = 8 amps
B) i = 8 (1-e^-1.5 t)
Vresistor = 6 i
you can put t = 5 in
Vcoil = 48 - Vresistor but check with
V = L di/dt :)
C) LOL, no di/dt at t=oo so no V across coil
The entire 48 volts is across resistor b then
1 answer