Pls answer to the question above
Attachments:
Answers
Answered by
1
Let the two junctions be labeled as C and D from left to right.
Between C and D, we have 3 resistances in parallel. They are (1+1+1)= 3 Ω on the top. 2Ω in the middle. Then (2+2+2)= 6Ω in the lowest branch.
So across CD : 1/R = 1/3 + 1/2 + 1/6 = 6/6
R = 1 Ω
A to B, the resistance is : series: 1 + 1 + 1 = 3 Ω
Between C and D, we have 3 resistances in parallel. They are (1+1+1)= 3 Ω on the top. 2Ω in the middle. Then (2+2+2)= 6Ω in the lowest branch.
So across CD : 1/R = 1/3 + 1/2 + 1/6 = 6/6
R = 1 Ω
A to B, the resistance is : series: 1 + 1 + 1 = 3 Ω
kvnmurty:
:-)
Thanks
click on the red heart thanks button above
Similar questions