Question

12.
Find the equivalent resistance across a and b in the circuit shown in the figure, if each resistance R with 8 resistances​

Answer 1

⭐《ANSWER》

$\huge\mathfrak{\underline {\underline\pink {ANSWER}}}$

↪Actually welcome to the concept of the CURRENT ELECTRICITY

↪Here , we basically know that , due to the same conducting wire around the resistors , there same potential difference around the Resistors

↪so we know that , due to same POTENTIAL DIFFERENCE , they are connect in parallel , so we get as

↪So finally while solving the simplified circuit we get as

〽Req = R/3 + R/3 + R

↪⭐=》 Req = 5R/3

Answer 2

Here,

let the resistances be named as,

R = R1 = R2 = R3 = R4 = R5 = R6 = R7

Now,

R1, R2 and R3 are in parallel combination.

Let their equivalent resistance be  R'

R5, R6 and R7 are in parallel combination.

Let their equivalent resistance be  R"

Now,

$\frac{1}{R'} = \frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}$

                        = $\frac{3}{R}$

R' = R/3

Similarly,

$\frac{1}{R"}= \frac{1}{R5}+\frac{1}{R6}+\frac{1}{R7}$

                        = $\frac{3}{R}$

R" = R/3

R' , R" and R4 are in series.

Thus their equivalence resistance be R"'

R"' = $\frac{R}{3}+R+\frac{R}{3}$

    = $\frac{5R}{3}$

Answer = 5R/3