Resistors connected in parallel and series

AI Thread Summary
To find the equivalent resistance between points A and B, the correct approach involves recognizing the configuration of the resistors. The 4 ohm and 3 ohm resistors are not in series; instead, the 3 ohm resistor is in parallel with the 2 ohm resistor. The equivalent resistance of the 3 ohm and 2 ohm resistors is 2 ohms, which is then in series with the first 4 ohm resistor, leading to a total equivalent resistance of 6 ohms. Misidentifying the connections can lead to incorrect calculations, emphasizing the importance of accurately visualizing the circuit layout. Understanding these principles is crucial for solving similar circuit problems effectively.
Cromptu
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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 :/


 

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Cromptu said:

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 :wink:
 
Cromptu said:
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...
 
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.
 
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