Question

Find the resistance of the circuit diagram

Right answer gets marked as brainliest

Answer 1

Answer:

8/3 Ω

Explanation:

As per the provided figure,

• R₁ = 1 Ω
• R₂ = 1 Ω
• R₃ = 1 Ω
• R₄ = 1 Ω
• R₅ = 1 Ω

Here, R₁ and R₂ are connected in series. So, combined resistance of R₁ and R₂ will be given by,

$\implies \sf {R_{(1,2)} = R_1 + R_2 }$

$\implies \sf {R_{(1,2)} = (1 +1) \; \Omega }$

$\implies \sf {R_{(1,2)} = 2 \; \Omega }$

Now, the combined resistance of R₁ and R₂ is connected with R₃ in parallel combination. So, combined resistance of R₁,R₂ and R₃ will be given by,

$\implies \sf {\dfrac{1}{R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} }$

$\implies \sf {\dfrac{1}{R_{(1,2,3)}} =\Bigg ( \dfrac{1}{2} + \dfrac{1}{1} \Bigg )\; \Omega }$

$\implies \sf {\dfrac{1}{R_{(1,2,3)}} =\Bigg ( \dfrac{1 + 2}{2} \Bigg ) \; \Omega }$

$\implies \sf {\dfrac{1}{R_{(1,2,3)}} =\Bigg ( \dfrac{3}{2} \Bigg ) \; \Omega }$

$\implies \sf {R_{(1,2,3)} =\dfrac{2}{3} \; \Omega }$

Now, the combined resistance of R₁,R₂,R₃ are connected in series combination with R₄ and R₅.

$\implies \sf {R_{(1,2,3,4,5)} = R_{(1,2,3)} + R_4 + + R_5 }$

$\implies \sf {R_{(1,2,3,4,5)} = \Bigg (\dfrac{2}{3} +1 + 1 \Bigg ) \; \Omega }$

$\implies \sf {R_{(1,2,3,4,5)} = \Bigg (\dfrac{2}{3} +2 \Bigg ) \; \Omega }$

$\implies \sf {R_{(1,2,3,4,5)} = \Bigg (\dfrac{2+6}{3} \Bigg ) \; \Omega }$

$\implies \sf {R_{(1,2,3,4,5)} = \dfrac{8}{3} \; \Omega }$

Therefore, the resultant resistance of the circuit is 8/3 Ω.