Battery and Current question

[Aviator4000]

I am totally lost on the battery and current questions. I have no idea what I am looking at (totally clueless). Can anyone help me out with this?

 

[Fly Navy]

I'm a EE. What do you need help on?

(Now let's see if I remember ANYTHING from my EE courses lol)

 

[Aviator4000]

FlyNavy:

Thanks for the response. How about an explanation of questions 9,10,11,31 on the ASTB gauge on this website. I would really appreciate it. Right now it might as well be in greek.

 

[jburnes]

9. The current in a loop like this isn't going to change directions, so the current will always be going counter-clockwise. If the current is at resistor "X" then the current will be flowing toward point "A". (answer is A)

10. Since the current is constant throughout the circuit because the resistors are in series, you can use V=IR to calculate the voltage drops. Just assume the current is equal to 1 Amp.
The voltage drop for the first resistor would be: 10 Ohm's x 1 Amp = 10 Volts
The voltage drop for the second resistor would be: 5 Ohm's x 1 Amp = 5 Volts
(answer is A)

11. Since this is a DC series circuit the current is constant because there is only one path for the electrons to flow and you can't just lose electrons. Current is equal to the # of electrons multiplied by their charge. So if the electrons are always travelling the same path, then the current will always stay the same. (answer is C)

31. In this problem circuit "A" is like that in problems 9,10 and 11. The current is the same everwhere in circuit "A", so V=IR looks like V=IR1 + IR2 and our total resistance is R = R1 + R2 (this is always the case for series ciruits, Rtot = R1 + R2 + R3 +...)

In circuit "B" the current takes two different paths, but the total current still remains the same. If V=IR then I=V/R. We want to look at each look independantly and then add them together.
The first loop contains the battery and R1. So the current for Loop1 is: I1=V / R1
The 2nd loop containf the battery and R2. So the current for Loop2 is: I2=V / R2

Now we can add the currents together to get: I = I1 + I2 = V/R1 + V/R2
if we divide by V we get: I/V = Rtot = 1/R1 + 1/R2 = 1/(R1+R2)

This is always the case for parallel ciruits: Rtot = 1/(R1 + R2 + R3 +...)

If we plug a few numbers in and let R1 = 4 and R2 = 6
Circuit "A" gives a total resistance of Rtot = 4 + 6 = 10
Circuit "B" gives a total resistance of Rtot = 1/(4+6) = 1/10 = 0.1

Thus circuit "A" has the greater resistance. (answer is A)

You're best bet to actually learn this stuff is to go to radio shack and buy a cheap multimeter a 9v battery and a pack of resistors, but if you have anymore questions I'll help you out.

 

[ea6bflyr]

Ah, you forgot the golden rule: if the resistance in each branch are equal, you can use:

RT = R/N;
where R = Resistance in each branch, and N is the Number of branches....
Example: if you have a parallel circuit with 2 branches with 20 ohms in each branch then the Total Resistance is equal to R/N = 20/2 = 10 ohms.

Hint, hint.....wink, wink...you might see this again!!

ea6bflyr

31. blah...blah...blah...
Now we can add the currents together to get: I = I1 + I2 = V/R1 + V/R2
if we divide by V we get: I/V = Rtot = 1/R1 + 1/R2 = 1/(R1+R2)

This is always the case for parallel ciruits: Rtot = 1/(R1 + R2 + R3 +...)

 

[jburnes]

ea6bflyr,

i should have read my post more carefully...i screwed up the math on the parallel circuit part.

Now we can add the currents together to get: I = I1 + I2 = V/R1 + V/R2
if we divide by V we get: I/V = 1 / Rtot = 1/R1 + 1/R2

This is always the case for parallel ciruits: 1/Rtot = 1/R1 +1/R2 + 1/R3 +...



I would have just edited my old post, but i can't find that button...

 

[Fly Navy]

Looks like I was beat to the punch.

 

[Aviator4000]

Thanks guys

 

[zuggerat]

electromagnetism --puke--