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Can someone help me with 2 circuit problems...please 
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#1
Jun2804, 12:47 PM

P: 47

1. The equivalent resistance between points A and B of the resistors shown below is 38 ohms. Find the value of the resistance R.
The resistor looks like this 55 Ohms _____\/\/\/\/_____   12 Ohms   A   .____/\/\/\__  ______. B     ____\/\/\/\/_____ R I have no idea what to do...please help. 2. Determine the magnitudes and directions of the currents through R1 and R2, where R1 = 36 ohms and R2 = 10 ohms. I already found the directions. they are both from right to the left, I just need the magnitudes. The Bs represent batteries, and the + next to them indicates the positive side of the battery. I am also really lost on this one can anyone help? V1=9.0V R1 +B/\/\/\    R2  ./\/\/\/.     B+ V2=6.0V 


#2
Jun2804, 03:27 PM

Sci Advisor
HW Helper
P: 2,586

I would suggest using [ code ] and [ / code ] tags:
SECOND EDIT: looks fine now  For the first question, you should know how to calculate net resistances when you have resistors in series and parallel. To find the total resistance between A and B, it is: [tex]R_{total} = 12 \Omega + (55 \Omega  R)[/tex] [tex]38 \Omega = 12 \Omega + (\frac{1}{55 \Omega} + \frac{1}{R})^{1}[/tex] [tex]26 \Omega = (\frac{1}{55 \Omega} + \frac{1}{R})^{1}[/tex] [tex](\frac{1}{26 \Omega}  \frac{1}{55 \Omega})^{1} = R[/tex] [tex]R \approx 49.310 \Omega[/tex] [tex]R = 49 \Omega[/tex]  Solve question 2 using the principle of superposition. That is, find the voltages and currents with the circuit first by replacing the top battery with a short circuit, then find the voltages and currents with the top battery, replacing the bottom with a short circuit. Add the numbers from the first calculations to the second calculations, and make sure you take direction/sign into account. 


#3
Jun2804, 06:34 PM

P: 47

Thank you so much, I got them both right and 100 on the assignment :) :) :)



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