Could u with this simple problem ?

[hsnelect]

Homework Statement

[PLAIN]http://img100.imageshack.us/img100/338/prob1o.jpg

Homework Equations

I = V / R

The Attempt at a Solution

3 + 5 = 8 A

 

[berkeman]

Homework Statement

[PLAIN]http://img100.imageshack.us/img100/338/prob1o.jpg

Homework Equations

I = V / R

The Attempt at a Solution

3 + 5 = 8 A

Welcome to the PF.

That looks correct for a)

How would you go about figuring out the rest of the questions?

 

[hsnelect]

Thanks Master!
That's really fast respond .. I'm impressed ^_^'

well .. I try to use the formula again I = V / R .. but it's not working .. :(

I heard that we need to use the Loop Method which I have no idea how to use! :(

I wonder how do we fine the equivalent V for the circuit! ??

 

[berkeman]

Thanks Master!
That's really fast respond .. I'm impressed ^_^'

well .. I try to use the formula again I = V / R .. but it's not working .. :(

I heard that we need to use the Loop Method which I have no idea how to use! :(

I wonder how do we fine the equivalent V for the circuit! ??

You have the resistances and currents for branches 1, 2, and 3. User V=IR to calculate the voltage drops across each resistor. Then the loop method just says that the sum of the voltage drops in any loop has to equal zero. You should be able to use this fact to calculate the two voltage sources.

 

[hsnelect]

Great ..
now I calculated the Voltage in each Resistor.
V @ R1 = 9v
V @ R2 = 32v
V @ R 3 = 45v

Now I need to add the sum of the voltage in the first loop and the second loop separately to get the 2 missing voltages ..

so..

9v + 32v + ε1 = 0

and

32v + 45v + ε2 = 0

is that correct ??

 

[berkeman]

Great ..
now I calculated the Voltage in each Resistor.
V @ R1 = 9v
V @ R2 = 32v
V @ R 3 = 45v

Now I need to add the sum of the voltage in the first loop and the second loop separately to get the 2 missing voltages ..

so..

9v + 32v + ε1 = 0

and

32v + 45v + ε2 = 0

is that correct ??

Yes, very good. Now the only thing to be careful about is the polarity of the voltage drops and voltage gains. Mark a +/- on the ends of the resistors, with the + end where the current enters the resistor (because current flowing through a resistor generates a voltage drop). The voltage sources would normally have their + end being the wide stripe end.

 

[hsnelect]

I got it .. ^_^ thanks!

The correct answer for both ε1 and ε2 is:

ε1 = 41v

ε2 = 77v

Great Job! ^_^

now how do we find the R on the top ..?? O_O

 

[berkeman]

I got it .. ^_^ thanks!

The correct answer for both ε1 and ε2 is:

ε1 = 41v

ε2 = 77v

Great Job! ^_^

now how do we find the R on the top ..?? O_O

Hint -- just keep using the V=IR equation...

 

[hsnelect]

imm .. yeah .. but the problem is .. this R experience two different voltages .. ! .. should I add them together or subtract or what should I do .. how do I know the true voltage on that resistor! :'(

Thanks! ^_^

 

[berkeman]

imm .. yeah .. but the problem is .. this R experience two different voltages .. ! .. should I add them together or subtract or what should I do .. how do I know the true voltage on that resistor! :'(

Thanks! ^_^

If the + terminals were pointing in the same direction for two power supplies in series, you would add the voltages. What should you do in this circuit?

 

[hsnelect]

WoW .. you are really physics master! ^_^

we subtract of course! ^_^

R = V/I

=> R = (77-41) / 2
=> R = 18 Ohms

Thanks! ^_^
I really like this Forum! ^_^