Solving series-parallel circuits

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SUMMARY

This discussion focuses on solving series-parallel circuits using the current and voltage divider rules. The total resistance calculated is 383Ω, with a total current of 0.03A and a current through R3 of 0.023A. The voltage across R1 and R2 is determined to be 2.97V, leading to a remaining voltage of 9.03V for the rest of the circuit. The voltage across R4 and R5 is calculated using the voltage divider rule, resulting in 1V for R4.

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LR5
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Summary:: Solving series-parallel circuits using current and voltage divider rule

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For the attached circuit, I need to determine the voltage drop across R4 using the voltage divider rule.
So far, I have determined total resistance to be R = 383Ω, total current I = 0.03A, and I3 = 0.023A.
I think these are correct.
I then get V through R1 and R2 to be 2.97V, therefore V for the rest of the circuit is 9.03V. Then the voltage for AB should be 9.03V.
Then V for R4 and R5 would be V ab (R45/(R45 + R3) = 9.03 x (41/371) = 1V.
I'm not sure if my methods and calculations are correct as this is very confusing :nb)
 
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LR5 said:
I'm not sure if my methods and calculations are correct as this is very confusing :nb)
How would we know if you are correct, as you show no work on getting the resistance? Show ALL the work used in reducing the circuit, step by step.
 

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