Answer:
a) The expression for electrical current: i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A
The expression for voltage: v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V
b) For t<=0 the inductor is storing energy and for t > 0 the inductor is delivering energy.
Explanation:
The question text is corrupted. I found the complete question on the web and it goes as follow:
The current in a 20 mH inductor is known to be: i = 40 mA at t<=0 and i = A1*e^(-10,000*t) + A2*e^(-40,000*t) A at t>0. The voltage across the inductor (passive sign convention) is -68 V at t = 0.
a. Find the numerical expressions for i and v for t>0.
b. Specify the time intervals when the inductor is storing energy and is delivering energy.
A inductor stores energy in the form of a magnetic field, it behaves in a way that oposes sudden changes in the electric current that flows through it, therefore at moment just after t = 0, that for convenience we'll call t = 0+, the current should be the same as t=0, so:
i = A1*e^(-10,000*(0)) + A2*e^(-40,000*(0))
40*10^(-3) = A1*e^(-10,000*0) + A2*e^(-40,000*0)
40*10^(-3) = (A1)*1 + (A2)*1
40*10^(-3) = A1 + A2
A1 + A2 = 40*10^(-3)
Since we have two variables (A1 and A2) we need another equation to be able to solve for both. For that reason we will use the voltage expression for a inductor, that is:
V = L*di/dt
We have the voltage drop across the inductor at t=0 and we know that the current at t=0 and the following moments after that should be equal, so we can use the current equation for t > 0 to find the derivative on that point, so:
di/dt = d(A1*e^(-10,000*t) + A2*e^(-40,000*t))/dt
di/dt = [d(-10,000*t)/dt]*A1*e^(-10,000*t) + [d(-40,000*t)/dt]*A2*e^(-40,000*t)
di/dt = -10,000*A1*e^(-10,000*t) -40,000*A2*e^(-40,000*t)
By applying t = 0 to this expression we have:
di/dt (at t = 0) = -10,000*A1*e^(-10,000*0) - 40,000*A2*e^(-40,000*0)
di/dt (at t = 0) = -10,000*A1*e^0 - 40,000*A2*e^0
di/dt (at t = 0) = -10,000*A1- 40,000*A2
We can now use the voltage equation for the inductor at t=0, that is:
v = L di/dt (at t=0)
68 = [20*10^(-3)]*(-10,000*A1 - 40,000*A2)
68 = -400*A1 -800*A2
-400*A1 - 800*A2 = 68
We now have a system with two equations and two variable, therefore we can solve it for both:
A1 + A2 = 40*10^(-3)
-400*A1 - 800*A2 = 68
Using the first equation we have:
A1 = 40*10^(-3) - A2
We can apply this to the second equation to solve for A2:
-400*[40*10^(-3) - A2] - 800*A2 = 68
-1.6 + 400*A2 - 800*A2 = 68
-1.6 -400*A2 = 68
-400*A2 = 68 + 1.6
A2 = 69.6/400 = 0.174
We use this value of A2 to calculate A1:
A1 = 40*10^(-3) - 0.174 = -0.134
Applying these values on the expression we have the equations for both the current and tension on the inductor:
i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A
v = [20*10^(-3)]*[-10,000*(-0.134)*e^(-10,000*t) -40,000*(0.174)*e^(-40,000*t)]
v = [20*10^(-3)]*[1340*e^(-10,000*t) - 6960*e^(-40,000*t)]
v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V
b) The question states that the current for the inductor at t > 0 is a exponential powered by negative numbers it is expected that its current will reach 0 at t = infinity. So, from t =0 to t = infinity the inductor is delivering energy. Since at time t = 0 the inductor already has a current flow of 40 mA and a voltage, we can assume it already had energy stored, therefore for t<0 it is storing energy.