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Kajayacht
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1. Find the equivalent resistance of the circuit as shown in the diagram below; where R1 = 1 Ω, R2 = 2 Ω, R3 = 2 Ω, R4 = 4 Ω, R5 = 1 Ω, R6 = 1 Ω, and R7 = 2 Ω.
[URL]http://img21.imageshack.us/i/prob06v2.gif/[/URL]
http://img21.imageshack.us/i/prob06v2.gif/
R(parallel) 1/R(total) = 1/R1 + 1/R2
R(series) R(total) = R1 + R2
I know I can do this one, its easy. But the arrangement of the resistors is throwing me off
I think I can take R3 and R4 in series, then take R5 in parallel with R3,4, take that in series with R2.
Then R7 and R6 are in series, and R7,6 is in parallel with R1. R1,7,6 in series with R2,5,3,4
Is this right?
[URL]http://img21.imageshack.us/i/prob06v2.gif/[/URL]
http://img21.imageshack.us/i/prob06v2.gif/
Homework Equations
R(parallel) 1/R(total) = 1/R1 + 1/R2
R(series) R(total) = R1 + R2
The Attempt at a Solution
I know I can do this one, its easy. But the arrangement of the resistors is throwing me off
I think I can take R3 and R4 in series, then take R5 in parallel with R3,4, take that in series with R2.
Then R7 and R6 are in series, and R7,6 is in parallel with R1. R1,7,6 in series with R2,5,3,4
Is this right?
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