Electricity Circuits?

Question

I have two very difficult physics electricity circuit questions that i have been trying to do for the past few hours. Can someone please try them and show your work so i can understand what you did to solve the questions.

R stands for Resistance!
V stands for Voltage!
I stands for Current!

The "Re" is the sum of all the resistance in the circuit.

They are posted on this webpage:
http://piczo.com/mohamadatwy?cr=1&rfm=y

Answer 1

Interesting web page!

Well, for starters YOUR Re is irrelevant; and your problem isn't a trivial one. It involves Kirchhoff's equations where the sum of the currents at a junction = zero.

But maybe this will get you started:

r3 and r4 are in series and their equivalent ohmage is 20.
That 20 ohm equivalent is in parallel with the 5 ohm r2, so the equivalent resistance of all 3 [r2, r3, and r4] is 4 ohms.

[ The equivalent resistance of resistances in parallel is the reciprocal of: {the sum of the reciprocals of each resistance}]

So now we have an equivalent 6+4+5 ohms load on the 30 volt batt. That means the current flowing thru r1 and r5 is 2 amps.

Now you have to "rebuild" your circuit, knowing that at the juncture of r1 with r2 and r3 you have 2 amps going in; and coming out, the greater part of that 2 amp current will pass thru the r2 (and not the series pair r3 & r4) while the remainder must pass thru the r3,r4 series pair.

As you can see, it's a rather tedious problem to type abt. Your second circuit is even worse.

A little more help: I will tell you that the voltage (measured at the ends of r2 is 8 volts. (You should have some idea how I got/get this.) That will give you an idea what current flows thru the r3, r4 series pair; and what current flows thru r2 (5ohm).

FURTHER, on your harder circuit the network resistance comes out very close to 10 ohms(9.994), unless I botched up some of the arithmetic; which means the 30 volt batt is feeding 3 amps out (and of course that's what current returns).

Answer 2

I only show the first one.

R3 + R4 = (12 + 8) ohm = 20 ohm
R2 // 20 ohm = 5 ohm // 20 ohm = 4 ohm
Re = R1 + R5 + 5 ohm = (6 + 5 + 4) ohm = 15 ohm
V = IR or
30 V = I x 15 ohm
I = 2 A

Th

Answer 3

Ok, I did some cheating for ya. Hey, you're cheating anyway, might as well get someone to help ya cheat, right? Ok, so here you go. Click the links below and they will show the circuits in question drawn in MultiSim2001 with voltmeters at every resistor showing the voltage at the respective resistor.

http://i83.photobucket.com/albums/j314/Jshwaa/answercircuit2.gif
http://i83.photobucket.com/albums/j314/Jshwaa/answercircuit3.gif
Granted I only gave you V. You can easily find I by taking V and dividing by the respective R. Good luck.