iain atkins

lol, ok Martin YHM then i guess.

I think its pretty straightforward stuff to those in the know.

www.race-tek.co.uk/photos/circuit question.doc

I've now hosted it if anyone can help it would be much appreciated.

Cheers



Iain

[Edited by iain atkins - 12/2/2003 1:36:57 PM]

Jiggerypokery

The value of R needs to be specified. After that, it's pretty simple.

The parallel resistance of 3 and 5 ohms together is R1:
R1 = [1 / ( (1/3) + (1/5) )]
The voltage V1 is :
V1 = 10 x R1 / (R1 + R)
Current A2 is :
A2 = 10 / (R1 + R)
Voltage between B anc C is:
10 - V1
Power comsumer will be:
I x I x R, which is:
A2 x A2 x (R1 + R)



[Edited by Jiggerypokery - 12/2/2003 3:17:31 PM]

dr_ming

Oh, I see, you're an anorak like me . That explains it . Mmmmm, Ohm's law, lovely....

[Edited by dr_ming - 12/2/2003 6:53:26 PM]

dr_ming

Kirchoff and Norton, now we're talking. Better add another voltage or current source though, just to make it worthwhile

[Edited by dr_ming - 12/2/2003 7:27:02 PM]

iain atkins

Basically its a circuit and he has to work out voltages at a certain point, amperes, and power consumed etc.

I think i can answer a couple for him.

But apart from that my physics / electronics is a bit rusty lol

Cheers



Iain

MartinM

Maybe....depends on the question!

Arron

Iain M8, you might be better off putting this in the Lighting and Other Electrical Section.

Just a thought old Buddy.

iain atkins

Nah geezer, it not to do with cars ~(for a change lol), its an exam question that i can't get my head around

iain atkins

um, engwish?

lol

Cheers m8

now in simpleton terms...... as i need to explain this later lol

Jiggerypokery

The resistance R1 is the result of connecting 3 ohms and 5 ohms in parallel.
The voltage across R1 is the input voltage (10V) multiplied by the proportion R1 is to the total resistance in the circuit.
The current flowing in the circuit is V (10V) divided by total resistance (R1 + R). From the equation V = IR.
The total voltage in the circuit is 10V, therefore V1 added to the voltage between B and C must add up to 10V.
The power consumption is P = I2 x R, where I is the current flowing in the circuit, and R is the TOTAL resitance of the circuit (R1 + R).

Still confused?

dr_ming

The voltage read by voltmeter V1 is given by the current through the 3 ohm resistor, which is 1A, multiplied by the resistance, 3 ohms. Therefore V1 = 1 x 3 = 3V.

Assuming the voltmeter V1 has infinite resistance, the current through the 5 ohm resistor is equal to the voltage across it (3V, from above), divided by the resistance, 5 ohms. Therefor the current through the 5 ohm resistor = 3/5 = 0.6A. The current through ammeter A2 is the sum of currents through the 3 ohm and 5 ohm resistors, therefore A2 = 0.6 + 1.0 = 1.6A.

Assuming ammeter A2 has zero resistance, the voltage between B and C is 10V minus the voltage across V1 (3V, from above). Therefore the voltage between B and C = 10 - 3 = 7V.

Power taken from the supply is given by the supply voltage multiplied by the total supply current. Total supply current is 1.6A from above, therefore power taken from the power supply is 10 x 1.6 = 16 watts (so I guess there is a miss-print in the question!).

An alternative way of calulating the power is to add up all the power dissipations in the resistors. For the 3 ohm resistor, power = 3V x 1A = 3W. For the 5 Ohm resistor, power = 3V x 0.6A = 1.8W. Power in resistor R = 7V * 1.6A = 11.2W (R = 7/1.6 = 4.375 ohms).

AndyC_772

iTrader: (2)

Jiggerypokery, you're missing something - there IS enough info here. Crucially, we know that there's 1 Amp flowing in the 3 Ohm resistor.

So, V1 reads 3V (V=IR, so V=1*3)
Ammeter A1 reads 0.6A (I=V/R, V=3 and R=5; the 5 Ohm resistor and 3 Ohm resistor are connected in parallel, so they see the same voltage).
Ammeter A2 reads 1.6A (summing currents at b; 1A plus 0.6A go in, so 1.6A must come out)
Voltage between b and c is 7V (just 10-3; if we call the node at the left of the diagram 'a', then Vac=10, Vab=3, so Vbc=7)
Power consumed is 16W - none of the answers offered is correct. This is simply V*I, where V is the 10V supply and I is the total current drawn, which is 1.6A.

Other interesting calculations:
R is 4.375 Ohms (R=V/I = 7/1.6)
3 Ohms in parallel with 5 Ohms gives 1.875 Ohms, giving a total circuit resistance of 6.25 Ohms.
This can then be used to confirm the power calculation; P=I^2*R = 1.6^2*6.25 = 16W. Also P=V^2/R = 100/6.25 = 16W.

HTH

Andy.

dr_ming

Andy, you must be nearly as sad as me.

AndyC_772

iTrader: (2)

*snigger*

Check my profile - let's just say I'm having a slow day at work

Jiggerypokery

Doh! You're right. I need my eyes tested.

hades

And there was me hoping for some nice, complicated nodal analysis.

CharlieWhiskey

You bunch of massochists

I should have been able to work it out, but couldnt be bothered

I will have to dig out some Kichoff questions to amuse you

CW
BEng(Hons) MIEE

iain atkins

Whoa.....how many replies lol

Cheers guys, much appreciated . I guess Andy / Dr Ming is correct then?

Thanks to you all for taking the time to help on this.

Cheers



Iain

[Edited by iain atkins - 12/3/2003 9:33:12 AM]

iain atkins

Andy / Dr

Oh and the power consumed bit, option c is actually 16w, not 16Kw as i had put.



Iain

civictyper

Yes, Dr Ming is correct.

I found the answers to be (albeit late):

3v
1.6A
7v
16W
R = 4.375 Ohms

So is this really for your mate ?

iain atkins

Nah, not for me lol.

Mate of mine seemed to think i would be able to work it out lol.