Quick question about simplifying a circuit

AI Thread Summary
The circuit reduction process involves combining resistors in series and parallel to simplify the overall resistance. The outer resistors, R5 and R6, are first combined in series to form R56. This is followed by combining R4 with R56 in parallel to get R456, and then adding R2 in series to find R2456. The next step involves combining R3 in parallel with R2456 to yield R32456, and finally adding R1 in series to determine the total resistance R132456. The discussion concludes that while the approach seems reasonable, there may be alternative methods to find the current i6.
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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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