Quick question about simplifying a circuit

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SUMMARY

The discussion focuses on the simplification of a circuit involving resistors R1, R2, R3, R4, R5, and R6. The user correctly identifies that the outer resistors R5 and R6 are in series, leading to the calculation of R56. The subsequent steps involve combining resistors in series and parallel configurations to derive the total resistance R132456. The user concludes that while the method appears sound, alternative approaches may exist for determining the current i6 more efficiently.

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Homework Statement



I want to see if I'm reducing the following circuit correctly.

Homework Equations





The Attempt at a Solution



The two outer resistors are in series:

##R_{56} = R_5 + R_6##

Now the outer resistors would be directly wired on each end after simplifying. Hence:

##R_{456} = (\frac{1}{R_4} + \frac{1}{R_{56}})^{-1}##

Series again:

##R_{2456} = R_2 + R_{456}##

Parallel:

##R_{32456} = (\frac{1}{R_3} + \frac{1}{R_{2456}})^{-1}##

Series:

##R_{132456} = R_1 + R_{32456}##

Now if we had the emf we could find the current and work backwards if needed.
 

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Looks reasonable. Looking for the current i6 there may be easier ways though. They should all amount to the same in the end.
 

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