So I need to find the total energy stored in this circuit at t = 0+, right after the switch changes position and at t -> infinity.
How would I go about doing that?
I poorly drew the circuit in paint. Here's a link to it:
http://dl.dropbox.com/u/1085106/circuit.…
Thanks in advance!
How would I go about doing that?
I poorly drew the circuit in paint. Here's a link to it:
http://dl.dropbox.com/u/1085106/circuit.…
Thanks in advance!
-
The only components that can store energy are caps, E = ½CV²
and inductors, E = ½LI²
Before the switch moves, calculate the voltage on the cap and the current in the L
The voltage across the cap is 6 volts so it's energy is ½(0.25)(6)² = 4.5 J
This is the same immediately after the switch changes as the cap voltage will not change immediately.
For the inductor, initially, the current is 1/2 of 5 amps, so the energy is ½LI² = ½(0.5)(2.5)² = 1.56 J. Note that the current divides into the two 4 ohm resistors.
And that is true also after the switch changes as the inductor resists changes in current.
After the switch has been closed for a long time, The voltage on the cap changes to the voltage that you would have on the 4 resistor near it. Wit 2.5 amps into to it, that is 10 volts. and the energy is ½(0.25)(10)² = 12.5 J
For the inductor, the current stays at 2.5 amps and the energy is ½LI² = ½(0.5)(2.5)² = 1.56 J
and inductors, E = ½LI²
Before the switch moves, calculate the voltage on the cap and the current in the L
The voltage across the cap is 6 volts so it's energy is ½(0.25)(6)² = 4.5 J
This is the same immediately after the switch changes as the cap voltage will not change immediately.
For the inductor, initially, the current is 1/2 of 5 amps, so the energy is ½LI² = ½(0.5)(2.5)² = 1.56 J. Note that the current divides into the two 4 ohm resistors.
And that is true also after the switch changes as the inductor resists changes in current.
After the switch has been closed for a long time, The voltage on the cap changes to the voltage that you would have on the 4 resistor near it. Wit 2.5 amps into to it, that is 10 volts. and the energy is ½(0.25)(10)² = 12.5 J
For the inductor, the current stays at 2.5 amps and the energy is ½LI² = ½(0.5)(2.5)² = 1.56 J