Question

Find the equivalent resistance in the following circuit ?

Answer 1

Answer:

$\huge\boxed{ \red{ \frac{11}{2} \Omega}}$

In this given circuit ,

• 2Ω and 2Ω are in parallel combination .

For this combination :

1/R'= 1/2+1/2=2/2=1

R' = 1Ω

• Also, 3Ω and 3Ω are in parallel combination.

For this combination:

1/R''= 1/3+1/3=2/3

R''= 3/2Ω

→ Now ,All becomes in series combination :

R(eq)= R'+R''+ 3Ω

→R(eq)= 1+3/2+3

→R(eq)=11/2 Ω

Suolution also is in attachment..

Answer 2

$\huge{\orange{\underline{\red{\mathscr{CASE.1}}}}}$

2Ω and 3Ω is parallel.

$\frac{1}{r'} = \frac{1}{2} + \frac{1}{2} \\ \frac{1}{r'} = \frac{\cancel{2}}{\cancel{2}} \\ \frac{1}{r'} = 1Ω.........(1)$

$\huge{\orange{\underline{\red{\mathscr{CASE.2}}}}}$

3Ω and 3Ω is parallel.

"

$\frac{1}{r"} = \frac{1}{3} + \frac{1}{3} \\ \frac{1}{r"} = \frac{2}{3} \\ r" = \frac{3}{2} Ω. \: \: ......(2)$

FROM 1 & 2.

R(eq)=1+3+3/2

R(eq)=2+6+3/2

.°.R(eq)=11/2Ω.

$\huge{\orange{\underline{\red{\mathscr{THANKS.}}}}}$