Cannot find V and I in this circuit

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AI Thread Summary
The discussion revolves around solving for current (i) and voltage (v) in a circuit using Ohm's Law and Kirchhoff's laws. The original poster struggled with the problem for 90 minutes, initially misinterpreting the units of Siemens (S) as resistance instead of conductance. After clarifying that conductances in parallel add while resistances in series add, they successfully converted the conductances to resistances. This led to the correct values of i = 6 Amps and v = 3 Volts. The conversation highlights the importance of unit understanding in circuit analysis.
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Homework Statement



Obtain the values of i and v in the circuit below:
http://img33.imageshack.us/img33/7538/newpicturenl.png
Answers: i = 6 Amps, v = 3 Volts

Homework Equations



Ohm's Law, Kirchhoff's Current Law & Kirchhoff's Voltage Law

The Attempt at a Solution



I have spent the last 90 minutes on this question!
I can see that it is simple, but I must be approaching it wrong as I keep on getting the same wrong answers.
I just need to know where to start...
Thanks
 
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id try combining the furthest resistances to the right because it is a series circuit, to get 9S then finding the total resistance of that and 2S which is 1/9S+1/2S=1/Rt ->2/18+9/18->11/18->18S/11= Rt. then find the total resistance of that and4s because its a series circuit. so its 18S/11 + 4S which is 62S/11... not sure if this is correct/and if it is idk where to go from there
 
You approached it the same way I did :)
I thought that way would work, but I kept ending up with i = 1.35 A
 
hmm, couldn't you do it that way then "simplify" it to a series circuit so I is constant. then can't you do V=IR and plug in 9 for Itotal and whatever the total resistance was for R and find V total? only problem here is S though.
 
The S unit is Siemens, which is conductance, not resistance. For an individual resistor, S = 1/R. This means that conductances in parallel add, while those in series add like resistances in parallel.

Sometimes it's easier just to convert all of the individual conductances to resistances...
 
Last edited:
I had no idea they were different units!
I thought the S was a variable...

Thank you for clearing that up for me - I've solved the question now, thanks. :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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