The Switch in the circuit shown is closed at time t=0.
|------------------8ohms-------5L (inductor)-------…
| |
|---------(Switch)-----27vbattery----…
1. At what rate is energy being disspiated as Joule heat in the resistor after an elapsed time equal to the time constant of the circuit? in units of W.
2. Calculate the rate at which energy is being stored in teh inductor at this time.
3. what is the total energy stored in the inductor at this time?
4. how long a time does it take the current to reach 89% of tis maximum value?
My Try:
v*(1 - e^(-1))
27V*0.63 = 17.07.
V^2/ R= P?
and now I'm lost
Please Help
|------------------8ohms-------5L (inductor)-------…
| |
|---------(Switch)-----27vbattery----…
1. At what rate is energy being disspiated as Joule heat in the resistor after an elapsed time equal to the time constant of the circuit? in units of W.
2. Calculate the rate at which energy is being stored in teh inductor at this time.
3. what is the total energy stored in the inductor at this time?
4. how long a time does it take the current to reach 89% of tis maximum value?
My Try:
v*(1 - e^(-1))
27V*0.63 = 17.07.
V^2/ R= P?
and now I'm lost
Please Help
-
Assume inductor is 5H (you've marked it 5L).
Time constant = L/R = 5/8 = 0.625 s
The final steady current (I_0) = V/R = 27/8 = 3.375A (because when the current is steady, the inductor has no effect - it behaves like a zero resistance).
Q1.
I=I_0(1-e^(-t/0.625))
When t = 0.625s, I = I_0(1-e^(-0.625/0.625))
= I_0(1 - e^-1) = I_0(1-0.368) = I_0 x 0.632 = 3.375 x 0.632 = 2.13A
Power in resistor = (I^2)R = 2.13^2x8 = 36.4W
Q2. Power supplied by battery = VI = 27x2.13 = 57.5W
Rate at which energy is stored in inductor = power from battery - power lost n resistor
=57.5 - 36.4 = 21.4W
Q3. Energy stored = LI^2/2 = 5 x 2.13^2 / 2 = 11.3J
Q4.
I=I_0(1-e^(-t/0.625))
0.89= 1-e^(-t/0.625)
e^(-t/0.625) = 1 - 0.89 = 0.11
Take natural logarithms of both sides
-t/0.625 = ln(0.11) = -2.21
t = 2.21 x 0.625 = 1.38s
Time constant = L/R = 5/8 = 0.625 s
The final steady current (I_0) = V/R = 27/8 = 3.375A (because when the current is steady, the inductor has no effect - it behaves like a zero resistance).
Q1.
I=I_0(1-e^(-t/0.625))
When t = 0.625s, I = I_0(1-e^(-0.625/0.625))
= I_0(1 - e^-1) = I_0(1-0.368) = I_0 x 0.632 = 3.375 x 0.632 = 2.13A
Power in resistor = (I^2)R = 2.13^2x8 = 36.4W
Q2. Power supplied by battery = VI = 27x2.13 = 57.5W
Rate at which energy is stored in inductor = power from battery - power lost n resistor
=57.5 - 36.4 = 21.4W
Q3. Energy stored = LI^2/2 = 5 x 2.13^2 / 2 = 11.3J
Q4.
I=I_0(1-e^(-t/0.625))
0.89= 1-e^(-t/0.625)
e^(-t/0.625) = 1 - 0.89 = 0.11
Take natural logarithms of both sides
-t/0.625 = ln(0.11) = -2.21
t = 2.21 x 0.625 = 1.38s