- #1
jg95ae
- 47
- 0
I'm not sure if I'm doing this right, I would appreciate if someone could take a look.
Q. Find the equivalent resistance of the circuit (attached diagram). Each of the six resistors has a resistance equal to 12 ohms.
I'm thinking that R5 and R6 are in series and therefore combine to be 2R.
2R is now in parallel with R4, therefore 1/Req = 1/R + 1/2R = 3/2R
Then 3/2R is in series with R2, so 2R/3 + R = 5R/3
Then 5R/3 is in parallel with R3, so 3/5R + 1/R = 8/5R
And since 8/5R is in series with R1, then 5R/8 + R = 13R/8.
Have I made any major blunders??
Q. Find the equivalent resistance of the circuit (attached diagram). Each of the six resistors has a resistance equal to 12 ohms.
I'm thinking that R5 and R6 are in series and therefore combine to be 2R.
2R is now in parallel with R4, therefore 1/Req = 1/R + 1/2R = 3/2R
Then 3/2R is in series with R2, so 2R/3 + R = 5R/3
Then 5R/3 is in parallel with R3, so 3/5R + 1/R = 8/5R
And since 8/5R is in series with R1, then 5R/8 + R = 13R/8.
Have I made any major blunders??