Help Required Please. This One Has A Constant Current Source!!

[kuiperbelt]

The question is to determine the current and its direction through the 3 ohm resistor.
See link:

https://ibb.co/dcVkbS4

[Hymie]

Use the same technique as advised in the circuit with 3 voltage sources.

Assign a current of 10A through the 12 ohm resistor (heading east); an arbitrary current (i1) though the 4 ohm resistor (heading south); an arbitrary current (i2) through the 3 ohm resistor (heading east); an arbitrary current (i3) through the 5 ohm resistor (heading south) and an arbitrary current (i4) through the 2 ohm resistor (heading west).

Based on ohms law, generate series of equations for the circuit loops (voltages/currents/resistances) and then use substitution to solve the equations.

[Atheist]

kuiperbelt Why are you bothered? These are just sums. Does it matter? I can see why, after a fashion. I remember learning about the way electricity flowed (not that it does flow!)
It's fun to work out the sums, but I have forgotten most of it. Thanks for reminding me of how it usd to be. Best wishes.

[kuiperbelt]

\kuiperbelt Why are you bothered?\

Around where I live people see you putting up a sensor security light or changing a brake light bulb on your car, and they think you are an expert LOL

Predominantly it is to help student neighbour on HND engineering course. Now it looks like my nephew is going to be asking for my help plus my grandson too! Oh no!! What have I let myself and all you AB users in for? LOL

These electrical problems are way above anything I encountered on my Physics degree, many many years ago!! I guess as well, one great benefit of being a member of AB, it is the cheapest solution to getting an education!! LOL

\learning about the way electricity flowed (not that it does flow!)\

Ahhh! Think I can help you there. I was taught that only current (amps) flows and a voltage is only ever present between two points. Does that answer your query?

[Etch]

If I've understood the circuit diagram correctly, I get 1.3559 Amps flowing from 'x' to 'y'.

[Old_Geezer]

Yeah, I recall this sort of stuff came up while doing an HNC. But I've more decades under my belt since then than before then; and forgotten how to tackle this sort of thing. Sad really, but there again, I don't recall ever using the knowledge.

[ZebuSanctuary]

Applying Kirchhoff's 1st law to the node at x;

-10 + Vx/4 + (Vx - Vy)/3 = 0

This reduces to ---> 7.Vx - 4.Vc = 120 --- Eq 1

Now apply K's 1st law to the node at y;

(Vy - Vx)/3 + Vy/5 + (Vy - 40)/2 = 0

This reduces to ---> 31.Vy - 10.Vx = 600 --- Eq 2

Eq 1 and Eq 2 form a simultaneous pair thus Vx = 34.576 Volts and Vy = 30.508 Volts.

Assuming conventional current flow through 3Ω resistor its value is (34.576 - 30.508)/3 = 1.356 Amps going eastbound.

https://ibb.co/sPKpMSk

[Etch]

Exactly the same method that I used, ZebuSanctuary.

[kuiperbelt]

Thanks to you all for your contributions.
Etch and Zebu bang on with your answers.I

BA to Zebu even though I think he has made a mistake in rearranging K's law for i1 in his diagram.

Thus -i1 = i2 + i3 Think you forgot the 'minus' sign ;-)

[ZebuSanctuary]

// made a mistake in rearranging K's law for i1 in his diagram //

Look again at the diagram. Observe the direction of the arrow representing i1.