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help me with this problem
- Thread starter groggory
- Start date
whoa, get a schematic editor... lol. if you have Visual Studio, it contains one (sorta), using one of the programs in there nobody ever uses... heh, can't think of the name. i use orcad pspice student edition, but i think it costs money... (i got it from school).
anyways, i'll take a look at the circuit later, i gotta run now though
one question though, Ix, is that a dependent current source? Where's the 2 Amp source?
anyways, i'll take a look at the circuit later, i gotta run now though
one question though, Ix, is that a dependent current source? Where's the 2 Amp source?
Originally posted by plot
whoa, get a schematic editor... lol. if you have Visual Studio, it contains one (sorta), using one of the programs in there nobody ever uses... heh, can't think of the name. i use orcad pspice student edition, but i think it costs money... (i got it from school).
anyways, i'll take a look at the circuit later, i gotta run now though
one question though, Ix, is that a dependent current source? Where's the 2 Amp source?
sorry, my editor is on my computer at home. I'm at work right now just working on my homework. So sorry about the crude drawing....
Ix is just showing the direction of current, as stated in drawing in the book. It is not any part of the circuit, only there for reference purposes.
The 2A source is that circle with an arrow in it. That is standard notation for an independent current source.
A dependent source looks like a diamond with an arrow in it with the dependent variable next to it. Something to the effect of ' 2Vx ' or something like that.
um.. well, havn't worked it out yet, but here's the simulation... might or might not help.
so, the voltage over the current source is -300, the current of course is 2 amps. P=I*V=-600watts.
to find the voltage over the current source... use superposition probably.
sry it's not much help, been awhile since i've had to deal with these hopefully this pushes you in the right direction however since it displays all values.
so, the voltage over the current source is -300, the current of course is 2 amps. P=I*V=-600watts.
to find the voltage over the current source... use superposition probably.
sry it's not much help, been awhile since i've had to deal with these
hopefully this pushes you in the right direction however since it displays all values.pretty straightforward... simplify the circuit to a series connection of a 100V voltage source, a 100 ohm resistor, and a 2A current source, like so:
Point "A" is 100V, obviously. The current through the resistor is set by the current source as 2A, which results in a voltage drop of 200V across the resistor - thus point "B" is 300 volts.
The current source is therefore providing 2 amps at 300 volts, which means it provides 600 watts, or "dissipates" -600 watts.
Making sure power is conserved... 2A @ 100V means the voltage source dissipates 200 watts, and I^2*R on the resistor gives a 400 watt dissipation. 600 - 400 - 200 = 0. yay!
Code:
A B
.--[100 ohms]--.
[+] (^) 2A
[-] 100V (|)
`-------.------'
VPoint "A" is 100V, obviously. The current through the resistor is set by the current source as 2A, which results in a voltage drop of 200V across the resistor - thus point "B" is 300 volts.
The current source is therefore providing 2 amps at 300 volts, which means it provides 600 watts, or "dissipates" -600 watts.
Making sure power is conserved... 2A @ 100V means the voltage source dissipates 200 watts, and I^2*R on the resistor gives a 400 watt dissipation. 600 - 400 - 200 = 0. yay!
awesome. Now I know the errors in my ways....
the way I was setting it up was going around the loop using KVL disregarding the current source as it has zero voltage drop across it. To get Ix. Then using KCL to on the point above the current law to get -2A. But I ended up multiplying by zero....yep, I did it wrong.
Thanks for setting me straight. I knew it was a simple problem, just didn't know how to go about doing it.
Thanks again.