I was wondering if i am doing this correctly,

if v1=30v than i can do R=V/I which becomes 30/10 = 3ohms correct?

please tell me if my calculations are correct or not and in case not, what i am doing wrong

thanks in advance

 

remember that, that 10A. current flow, is going to divide between the parrallel combination of R and 3R.

current through R will equal (10A. x 3R / (R + 3R)

 

i am kind of stuck now, i do know how to proceed, so how can i calculate R?

 

for starters,
you have a voltage drop across the 12 ohm resistor.

 

which is -120v, i got that part, and than i know the Ir+I3r has to = 10A. i have the puzzle parts i just having trouble putting them together but i appreciate all of your help that you been giving me

 

I don't have it all figured out myself either, I'm just working with it too, in between comercials.

 

i feel better that its just not me, there is always a piece missing for me

 

are they giving the voltage v1 as 30 volts. or are you just using that as an example, for clarity?

 

they are giving it at 30V

 

good that helps out, in calculations.

 

throwing this out quickly,
10A. ci\oming in,
wouldn't 1/3 flow through 3R?

So asume R=1 ohm

 

thats what i am thinking but i dont know if the 3 means that or its just their for being their

 

It means 3 x R

 

so the current on r has to be 3 times as 3r?

 

I assume the 30 volts is supplied in part a of your question?

So we have 120 volts across the 12 ohm resistor and 30 volts across the parallel combination of R and 3R. Call this Rp

So the 120 + 30 = 150 volts is distributed across the voltage divider formed by 12 and Rp
such that 30/150 = Rp/(12+Rp)

So you can calculate Rp

So you can work back to R from the parallel resistance formula

In part b you don't know the voltage across R and 3R or the currents in them, but you do know the power dissipated by their combination.

Power in R power in 3R = 1575 - power in 12ohm resistor = 1575 - (10*10*12)

This power is distributed between the two resistors. You must calculate the ratio of the currents to obtain the ratio of the powers.

 

but 30v is just through r and not 3r so that doesnt work i believe

 

30v. is across R and 3R.

R and 3R form a parrallel combo, that acts as a single resistance in series with the current.
But than when the current reaches this combo. it divides.

 

You have put up a circuit including a current source.

It is my assumption that you have a working knowledge of series and parallel resistors, since you will have covered these maybe 2 years before current sources are introduced.

If that is not the case please review series and parallel resistors and the formulae involved.

When you do this you will find that the voltage across every resistor in a parallel combination is the same. However the currents are different and depend upon their relative values.

You will also find that the current through every resistor in a series combination is the same. However the voltages are different and depend upon their relative values.

 

yes you are right its just been a long day looking at this

 

I assume the 30 volts is supplied in part a of your question?

So we have 120 volts across the 12 ohm resistor and 30 volts across the parallel combination of R and 3R. Call this Rp

So the 120 + 30 = 150 volts is distributed across the voltage divider formed by 12 and Rp
such that 30/150 = Rp/(12+Rp)

So you can calculate Rp

So you can work back to R from the parallel resistance formula

In part b you don't know the voltage across R and 3R or the currents in them, but you do know the power dissipated by their combination.

Power in R power in 3R = 1575 - power in 12ohm resistor = 1575 - (10*10*12)

This power is distributed between the two resistors. You must calculate the ratio of the currents to obtain the ratio of the powers.

the 120v is a voltage drop, i was calculating the 12Ω x 10a so we shouldnt add that to the 30 right? the only info supplied is what is on the drawing and that v1=30