Resistors And Battery Diagram Poser

[kuiperbelt]

The son of our next door neighbour is doing HND in Electrical and Electronic Engineering and requires some assistance.
He has asked me to help but it's a little beyond me. I told him that I'm sure someone on AB can give him a pointer if not how to arrive at the answer :-)
The question is to find the current through the 5 ohm resistor and the voltage across the top of the 2 ohm resistor and the bottom of the 20 V battery. He does have the answers but needs a method to solve the problem!! See link below:

https://ibb.co/ggqzkhF

[The Builder]

Put an Ammeter in series with the 5ohm resistor to measure the current going through it.

Put a Voltmeter between the top of the 2 ohm resistor and the bottom of the battery to measure the voltage between them.

It's been a while, but I think that'll do it ;o)

[Togo]

V
I = --
R

I
V = --
R

V
R = --
I

[Togo]

Bleedin site edited the layout. He needs to know those 3 formulas like the back of his hands and understand them. Doing the number crunching for him will not help him later on because he will not instinctively spot a faults in circuits that he may need to repair.

I = V/R

V =I/R

R = V/I

[bobbinwales]

Yes thems the formula but you also need to know how to calculate when you got resistors in sequence and in parallel

[Etch]

Second of Togo's formulae should be
V = I x R

[Togo]

V ...................................................................................................
I = --
R.....................................................................................................

I ....................................................................................................
V = --
R ...................................................................................................

V ..................................................................................................
R = --
I ....................................................................................................

[Togo]

Give up with this site's answer box characteristics.

[Togo]

It should Etch. V =I x R

It has been a while. :))

[The Builder]

Sorry. I didn't realise it meant for you to "calculate" the results. :o(

[Togo]

True again Bobbin. But if he does not have those basic formulae in his locker then series and parallel circuitry for resistors, or indeed any other electronic device mean nothing. He must have covered these principles or he would not be getting the question asked. It could be a matter of life and death later, not just his own, if he were to qualify in that field of expertise and not deserve to have done so.

[Etch]

The voltage/current/resistance relationship is known as Ohm's Law:
https://en.wikipedia.org/wiki/Ohm%27s_law#Circuit_analysis

[Togo]

The young man will note that there is a difference of 10v across the 3 ohm and 5 ohm resistors in series. He must also reckon on the fact that there is also a 15v battery with a 1 ohm resistor between itself and ground(important) in the circuit. First thing I would do is find the total current flow across the circuit from ground to 25v assuming that a battery has no resistance or impedance( it does but it will not matter in that circuit) A clue. batteries in parallel of 15v and 20v do not total 35v. In series they do.

[Hymie]

No one has really given the answer of how to solve this problem; Ohms law on its own is not enough.

To solve the problem, arbitrarily assign current values through each of the circuit current paths (the direction is somewhat unimportant as the result + or – will indicate the actual current flow direction).

If the current through the 3R resistor is assigned i1 (heading east), and the current through the 2R resistor (heading north) i2, then current through the 5R resistor (heading east) will be i1 + i2.

Then using Ohms law (based on the voltages differences and resistances) create equations (using the above assigned currents) for the potential different paths in the circuit, and then solve the equations by substitution to find the currents (i1 & i2).

Once the currents are known, the voltages at all points in the circuit can be found by Ohms law.

[johnk]

Is that HND standard nowadays!?

[bobbinwales]

Not sure I ever said this before but I agree with Hymie....

[Hymie]

To assist with your understanding, here is a worked example with two voltages souces, but the principles are the same:-

Power Dissipated in Load Resistor with Two Voltage Sources - YouTube

[Hymie]

try again

[ZebuSanctuary]

Kuiper - Just to make sure we are on the same page.

You have said the HND student possesses the answers.

Does the current flowing through the 5Ω resistor equal 0.972 Amps?

If it does, then I'll provide an explanation for the student using Kirchhoff's second law.

Awaiting your reply.

[Etch]

I also get 0.972 A for the current through the 5Ω resistor, and 20.833 V for the voltage across the 2Ω resistor and 20 V battery.

[kuiperbelt]

Zebu, Etch \0.972 Amps\

Yes, exactly! The other answer is 20.833 volts!

My neighbour's son says as well as Kirchhoff's, it can be solved by using the Thevenin Theorem or the Superposition theorem too.

I look forward to your explanations - Please!!