The circuit shown below has a 100 Ohm resistor and 50 Volt-sec/amp (a unit of inductance called the Henry) inductor connected to a 12V battery. It is a freshman physics tradition that inductors are very poorly The switch is closed at t = 0 seconds. What is the magnitude of the current in amps at t = 1 second?drawn.
V = i R + L di/dt
i will be of form i = I (1-e^kt)
where I is the current after a long time determined only by the battery and the resistance because the inductor has no resistance once steady stage is reached.
at t = infinity, i =V/R
so I = V/R
and at t = 0 , V = L di/dt and di/dt=-kI
so V/L = -k V/R and therefore k = -R/L
so
i = V/R (1-e^-Rt/L)
i = (12/100)(1-e^-100t/50)
= .12(1 - e^-2t)
if t = 1 then
i = .12 (1- e^-2 )
e^-2 = .1353
so
i = .12 (1 - .1353)
= .104 amps
di/dt = -Ike^-kt
V = i R + L Ik(-e^-kt)
V = IR(1-e^kt) - LIk e^-kt
V = I R -I (R + Lk)e^-kt
at t = 0, V = I R
so R + Lk = 0
I = V/R =12/100 = .12 amps
forget the last few lines there, you do not need it
To determine the magnitude of the current in the circuit at t = 1 second, we need to use the concepts of inductance and the properties of an inductor.
First, let's consider the basic equation of an inductor:
V = L * di/dt
where V represents the voltage across the inductor, L is the inductance, and di/dt is the rate of change of current through the inductor with respect to time.
In this case, the inductance is given as 50 Volt-sec/amp or 50 H (Henry), and the battery voltage is 12V. Therefore, we can write the equation as:
12V = 50H * di/dt
Now, we need to solve this differential equation. To do that, we can rearrange the equation as:
di = (12V / 50H) * dt
Integrating both sides of the equation, we have:
∫di = ∫(12V / 50H) * dt
∫di = (12V / 50H) ∫dt
i = (12V / 50H) * t + C
Here, C is the constant of integration, which we need to determine using the given initial condition at t = 0 seconds. Since the switch is closed at t = 0 seconds, initially there is no current flowing through the inductor, i.e., i(0) = 0. Substituting this condition into the equation, we have:
0 = (12V / 50H) * 0 + C
C = 0
Now we can go back to our equation:
i = (12V / 50H) * t
Substituting t = 1 second into this equation:
i(1) = (12V / 50H) * 1
i(1) = 0.24 A (rounded to two decimal places)
Therefore, the magnitude of the current in amps at t = 1 second is approximately 0.24 A.