Circuits - Calculating Effectice Resistance

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The discussion focuses on calculating effective resistance in circuits that contain both series and parallel components. For part A, the user calculated the total resistance by first summing the series resistances and then the parallel resistances, arriving at a total of 5Ω. In part B, the user initially struggled but then broke down the circuit into manageable branches, correctly identifying the 12, 4, and 16 Ohm resistors as being in parallel. After calculating the effective resistances for each branch, they found the overall effective resistance to be 3.88Ω. The conversation highlights the importance of simplifying complex circuits into smaller parts for easier calculations.
FaraDazed
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Sorry for my crude drawings but I thought it was easiest and quickest this way to post the question.

Homework Statement



Calculate the effective resistance for the circuits below.

A.
circuit_A.jpg


B:
circuit_B.jpg



Homework Equations


R_{series}=\sum\limits_{i=1}^n R_i \\<br /> \frac{1}{R_{parallel}}=\sum\limits_{i=1}^n \frac{1}{R_i}<br />


The Attempt at a Solution



We have never been told how to calculate the effective resistance where both series and parallel occur, only how to do it if it is series and how to do it if it is parallel. But using intuition for part A i just did the bit in series and then the bit in parallel and added them like below.

<br /> \sum\limits_{i=1}^n R_i = 2+2=4 \\<br /> \sum\limits_{i=1}^n \frac{1}{R_i} = \frac{1}{2}+\frac{1}{2} = 1 ∴ R=1 \\<br /> ∴R_{total}=4+1=5Ω<br />

For part B I am scratching my head as to where to start because the whole thing is in parallel and then there are parrellel bits inside of that iyswim. Any hints on how to approach part B would be much appreciated, thanks.
 
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Forget about the rest of the circuit .First consider only 12,4,16 Ohm resistors in the top branch .What do you think ? Are they in series or parallel ? Can you calculate their effective resistance ?
 
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Tanya Sharma said:
Forget about the rest of the circuit .First consider only 12,4,16 Ohm resistors in the top branch .What do you think ? Are they in series or parallel ? Can you calculate their effective resistance ?

Right ok, I think I see what your getting at. So as the 12,4 and 16 are in parrallel I can calculate the effective resistance of just that, ignoring the rest (for now), and replace the 12,4,16 branches with the effective resistance. Ok I will have a go now.
 
Yes...that is correct .
 
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OK so for part B then I broke it down to the individual branches to make it easier on the eye.

Top Branch:
<br /> 3+9+(\frac{1}{12}+\frac{1}{4}+\frac{1}{16})^{-1}=14.53Ω

Middle Branch<br /> 20+10=30Ω

Bottom Branch<br /> 5+(\frac{1}{8}+\frac{1}{33})^{-1}=6.44Ω

Putting it all together<br /> (\frac{1}{14.53}+\frac{1}{30}+\frac{1}{6.44})^{-1}=3.88Ω<br />
 
Nice work :thumbs:
 
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Thanks for your help :)
 
You are welcome :smile:
 
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