Using Current and Voltage Dividers

[rugerts]

Homework Statement

Find:
A) vo
B) io
C) power dissipated in 6 ohm resistor
D) power developed by current source

Relevant Equations

Current and Voltage divider equations. Ohm's law.

Can anyone explain how to go about this? I don't understand the solution, which says:
A) Find the equivalent resistance (10 ohms). Find the voltage of the current source (4.2*10 = 42 V). Use a voltage divider to find vo = 42*(20/(10+20))=28V. My question is why do we need to add 10, which is R equivalent to 20 again? I thought I could just put 10 in the denominator, since this is R equivalent.
B) Similar questions. Voltage across 90 ohm resistor = 42*((90 in parallel with 10)/(6+(90 in parallel with 10)) = 25.2 V. Current across 90 ohm resistor = 25.2/90 = 0.28 A.
C) Power dissipated by 6 ohm resistor = (Voltage across current source - voltage across 90 ohm resistor)^2 / 6. My question here is how do I know which is positive and which is negative here in terms of subtraction/addition.

 

[kuruman]

Please post the full statement of the problem before seeking help. We cannot help you without it. For example, where does the 42 come from?

 

[rugerts]

Please post the full statement of the problem before seeking help. We cannot help you without it. For example, where does the 42 come from?

sorry, the only thing I left out was that ig, the current source, is 4.2 A

 

[kuruman]

I wouldn't worry about voltage divider equations. You have a total current of 4.2 A that is split between the branch on the left and the branch on the right and two equivalent resistances on the the left and on the right. If you find the equivalent resistances, you can find what fraction of the total current flows in each branch. Once you have that, the rest should be easy using Ohm's law.

To answer your specific question for (a), note that the current in the $20~\mathrm{\Omega}$ and the $10~\mathrm{\Omega}$ resistors is the same. You know that $V=IR$ for each resistor. This means that the $20~\mathrm{\Omega}$ resistor has twice the voltage across it than the $10~\mathrm{\Omega}$ resistor. How do you split 42 V into two pieces such that one is twice as large as the other? Answer: You split it into three pieces and give two to the larger resistor and one to the smaller resistor, $\frac{2}{3}=\frac{20}{20+10}$.

You don't need to know any signs for (c). $P=I^2R$. If you find the current that flows in the circuit on the right as I suggested earlier, you are in business.