Voltage across resistors - Series & Parallel

[gneill]

I just spotted where in post#23 you multiplied the current by the ENTIRE RESISTANCE of the string of resistors rather than the equivalent resistance R234. You want the voltage across R234, so you must multiply the current by R234.

 

[gneill]

No there's nothing wrong with that at all but I still don't understand how I find V1.

If the Battery Voltage is 10V and the Current Flowing in the circuit is 9.578mA and the circuit resistance 1.044KΩ, I don't understand what I have to do with these values to find V1, V2 & V3.

Ohm's Law! Current x Resistance = Voltage. You have the current through R1, R2334, and R5. Multiply each of those resistances by its current and you get the potential drop across each resistance, i.e., V1, V2, and V3.

 

[lloydowen]

I just spotted where in post#23 you multiplied the current by the ENTIRE RESISTANCE of the string of resistors rather than the equivalent resistance R234. You want the voltage across R234, so you must multiply the current by R234.

So 330Ω+470Ω+560Ω = 1360Ω

1360Ω * Battery current = 9.578mA

Which gives 13026.08V

WHAT!?

 

[lloydowen]

Or should I convert the 9.578mA into amps so its 0.009578A?

 

[lloydowen]

OMG I just clicked! Thanks! :) Give me a few minutes to write this down I may have another question Thanks so much so much more help than my tutor!

 

[technician]

The resistors 330, 470 and 560 are in parallel... you do not simply add these to get the combined resistance.
A little tip... the combined resistance of a parallel combination MUST be LESS THAN the value of any resistor in the parallel combination... the combined resistance must be less than 330

 

[lloydowen]

The resistors 330, 470 and 560 are in parallel... you do not simply add these to get the combined resistance.
A little tip... the combined resistance of a parallel combination MUST be LESS THAN the value of any resistor in the parallel combination... the combined resistance must be less than 330

So then do I do R234 Rt = 1/330 + 1/470 + 1/560 ?

 

[lloydowen]

Meaning 144Ohms x 9.578mA which gives me 1.379V which is what PCbwizard calculated :) :)

 

[lloydowen]

So how now do I go about finding the currents I2 - I4 ?

 

[technician]

You have got it... can you now calculate the current in each of the parallel resistors
You know the voltage across the parallel combination... it is the same for each resistor... just use V = I x R
You should find that the three currents add up to 9.578mA

 

[lloydowen]

So now I've done V2 (1.379V)/R2(330Ω)

which gave me 4.18mA... does that sound reasonable ?

 

[technician]

You have done it... do exactly the same for the 470 and 560 resistors
You should get the 3 currents are 4.18mA, 2.9mA and 2.4mA which add to 9.48mA (near enough with rounding off)
Well done

 

[lloydowen]

I have done it! :) Thanks guys so much help! I've actually learned something :P Is there a way I can say thanks on this forum like a +1 rep or?

 

[gneill]

I have done it! :) Thanks guys so much help! I've actually learned something :P Is there a way I can say thanks on this forum like a +1 rep or?

Well done lloydowen. The best way to thank us is to ace your next exam! If you have comments you wish to make about your experience here, there's always the "Forum Feedback & Announcements" forum in the PF Lounge area.

 

[lloydowen]

I will be sure to write in there! I'm taking Electrical Engineering and I have an exam monday so I'll hopefully be good for it! :P

 

[technician]

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