Resistors connected in parallel and series

[Cromptu]

Homework Statement

Find the equivalent resistance between the points A and B of the network shown in figure ( file has been attached )

Homework Equations

Req( series) = R1+ R2 + R3 .. + Rn
1/Req ( parallel) = 1/R1 + 1/R2 + 1/R3 .. + 1/Rn

The Attempt at a Solution

We have to find the equivalent resistance between a and b.
The way I solved it was :
1) the second ( from left )4 ohm and the 3 ohm are connected in series, so Req = 7 ohm.
2) now the 7 ohm and 2 ohm are connected in parallel, so Req = 1.56 ohm
3) the remaining 4 ohm and 1.56 are connected in series, so Req = 4.56 ohm.

But this answer is wrong..According to the book,
The 2 ohm resistance is connected in parallel with the 3 ohm resistor. Their equivalent resistance is 2 ohm which is connected in series with the first 4 ohm resistor from left. Thus equivalent resistance between A and B is 6 ohm.
Please help :/

 

[gneill]

Homework Statement

Find the equivalent resistance between the points A and B of the network shown in figure ( file has been attached )

Homework Equations

Req( series) = R1+ R2 + R3 .. + Rn
1/Req ( parallel) = 1/R1 + 1/R2 + 1/R3 .. + 1/Rn

The Attempt at a Solution

We have to find the equivalent resistance between a and b.
The way I solved it was :
1) the second ( from left )4 ohm and the 3 ohm are connected in series, so Req = 7 ohm.

No, those resistors are not connected in series; The first 4 Ω resistor also connects at the node where they meet. For two components to be in series, no other component can share their connection point.

I suggest that you begin from the 'back end" of the circuit, with the 2 Ω resistor, and work "forward" from there

 

[pcm]

1) the second ( from left )4 ohm and the 3 ohm are connected in series, so Req = 7 ohm.

4 and 3 are not in series.think...

 

[godwhite]

the problem with these questions is that, you think that some of the resistors are in series with the other, where in reality they are not. Instead of drawing the R3 resistor on the same horizontal line as the other 4 ohm resistor(R1), you could move it closer to the 2ohm resistor and draw them on the same vertical line.

You will see that R3 and R4 are actually in series, which means you could add it up together, which gives 6ohms. We'll name this equivalent resistor R34. You will also notice that R34 is parallel to R2. So 1/(1/6 + 1/3) = 2 ohms. Now we name this simplified resistances R234.

R234 is in series with R1. 2+4 = 6ohms.

 

[asteriatic]

Your approach to solving this problem is correct, but it seems like there may have been a mistake in your calculations. When finding the equivalent resistance of resistors in parallel, you need to take the reciprocal of each resistor and then add them together before taking the reciprocal again. So for the first step, the equivalent resistance of the 4 ohm and 3 ohm resistors in series would be 1/(1/4 + 1/3) = 12/7 ohm. Then, for the second step, the equivalent resistance of the 7 ohm and 2 ohm resistors in parallel would be 1/(1/12 + 1/2) = 24/13 ohm. Finally, for the third step, the equivalent resistance of the remaining 4 ohm and 24/13 ohm resistors in series would be 4 + 24/13 = 76/13 ohm. This is not equal to the answer given in the book, so it's possible that there is a typo or mistake in the book's answer. I would recommend double checking your calculations and also consulting with your teacher or a classmate to see if they got a different answer.