Question

Q. 3

Plz answer it asap​

Answer 1

Answer:

2Ω

Explanation:

The upper resistances of each  3Ω are in series, so

$R_{eq}= R_{1}+R_{2}=3+3=6$Ω

Similarly, the resistances below are also in series,

so,

$R_{eq}= R_{3}+R_{4}=3+3=6$Ω

Now, the resultant resistances at the top and bottom and the center are in parallel, and equal also, so

$R_{eq}=\frac{R(\;the\;resistances\;in\;parallel)}{n\;\;(no.\;of\;resistances)}\\=> R_{eq}=\frac{6}{3}=2 oh,$

$Learning\;\;together\;:)$