Circuit problem -- Voltage source and 5 resistors....

[terryds]

Homework Statement

http://www.sumopaint.com/images/temp/xzkaelnnkibopbdq.png

The electric current that goes through the 5 ohm resistor is

A. 0 A
B. 0,3 A
C. 0,6 A
D. 0,9 A
E. 1,2 A

Homework Equations

1/R_parallel = 1/R₁+1/R₂+...
R_series = R₁+R₂+...

The Attempt at a Solution

I take the 6 ohm and 3 ohm as series so it's 9 ohm
Then, 4 ohm and 2 ohm as series so it's 6 ohm
Then, I take the 9 ohm, 6 ohm, and 5 ohm as parallel

But, it gives me 2,09 A which doesn't exist in the option..
Please help me

 

[rude man]

I take the 6 ohm and 3 ohm as series so it's 9 ohm

Can't do that. There is the + side of the battery inbetween.
Mark v1 and v2 as the voltages across the 5 ohm respectively. Call the - side of the battery 0V.
Then write KVL to solve for v1 - v2.

 

[terryds]

Can't do that. There is the + side of the battery inbetween.
Mark v1 and v2 as the voltages across the 5 ohm respectively. Call the - side of the battery 0V.
Then write KVL to solve for v1 - v2.

Hmm... I use KVL, and get the current in the top closed loop is 0,3 A.
And, the bottom closed loop is 0,6 A.
So, what's the current that goes through the 5 ohm resistor?
I think it must be 0,9 A, right? Since the current at the top will meet the current at the bottom (the current at the top goes down, and the current at the bottom goes up)
I thought of using KVL at the closed loop in the 4ohm,2ohm,5ohm loop. But, it will be all variables and no numbers (since -5I - 4I + 2I = 0), and I'm not sure if I have had the sign (positive/negative) right...

 

[rude man]

Hmm... I use KVL, and get the current in the top closed loop is 0,3 A.
And, the bottom closed loop is 0,6 A.
So, what's the current that goes through the 5 ohm resistor?
I think it must be 0,9 A, right? Since the current at the top will meet the current at the bottom (the current at the top goes down, and the current at the bottom goes up)
I thought of using KVL at the closed loop in the 4ohm,2ohm,5ohm loop. But, it will be all variables and no numbers (since -5I - 4I + 2I = 0), and I'm not sure if I have had the sign (positive/negative) right...

I was expecting you'd write 2 equations in 2 unknowns: v1 and v2. Then current thru 5 ohm is (v1 - v2)/5.

 

[terryds]

I was expecting you'd write 2 equations in 2 unknowns: v1 and v2. Then current thru 5 ohm is (v1 - v2)/5.

How to determine v1 and v2?
I have no idea. Please give me some explanation on how to get v1 and v2

 

[gneill]

How to determine v1 and v2?
I have no idea. Please give me some explanation on how to get v1 and v2

It would be helpful if we had an idea of what circuit analysis techniques you've studied so far. For example, have you covered any of these: Nodal Analysis, Mesh Analysis, Thevenin Equivalents

 

[terryds]

It would be helpful if we had an idea of what circuit analysis techniques you've studied so far. For example, have you covered any of these: Nodal Analysis, Mesh Analysis, Thevenin Equivalents

I've just studied Ohm's law and Kirchchoff's law so far...
Could you please give and explain me the answer?
Thanks anyway

 

[gneill]

I've just studied Ohm's law and Kirchchoff's law so far...
Could you please give and explain me the answer?
Thanks anyway

Sorry, we can't do your homework for you here, just give advice.

If you're limited to the fundamentals of Ohms Law, KVL, and KCL, then you're in for a bit of work. You'll have to assign currents to each resistor and write KCL for each node, KVL for each loop, and solve for the currents.

On the other hand, if you're allowed to be a bit sneaky and take advantage of symmetry you can get there much quicker. Note that the resistors in the two left loops have the same ratios: 6:4 is the same as 3:2. Does that tell you anything about what potentials will end up on the top and bottom nodes? Hint: Suppose the 5 Ω resistor were removed temporarily. Taking the battery negative terminal as a reference node, what will V1 and V2 be in the figure here:

 

[terryds]

It seems like a balanced wheatstone bridge. So, v2-v1 is zero, right ?? So, the current is 0 ampere ??

v1 = 4/10 * 3 = 1,2
v2 = 2/5 * 3 = 1,2

v2 - v1 = 0

Is it right ??

 

[gneill]

Correct.