Question

Three resistors of 1 Ω, 2Ω and 3 Ω are connected in parallel. The combined resistance of the three resistors should be​

Answer 1

Answer:

The equivalent resistance of the given combination is 6/11 Ω

Step-by-step explanation:

• When resistors of resistances R₁ , R₂ , R₃ .... are connected in parallel, the equivalent resistance (R) is given by

  1/R = 1/R₁ + 1/R₂ + 1/R₃ + ...

Let

• R₁ = 1 Ω

• R₂ = 2 Ω
• R₃ = 3 Ω

Substituting the values,

$\sf \dfrac{1}{R}=\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3} \\\\ \sf \dfrac{1}{R}=1+\bigg(\dfrac{1 \times 3}{2 \times 3}+\dfrac{1 \times 2}{3 \times 2} \bigg) \\\\ \sf \dfrac{1}{R}=1+\dfrac{3}{6}+\dfrac{2}{6} \\\\ \sf \dfrac{1}{R}=1+\dfrac{3+2}{6} \\\\ \sf \dfrac{1}{R}=1+\dfrac{5}{6} \\\\ \sf \dfrac{1}{R}=\dfrac{6+5}{6} \\\\ \sf \dfrac{1}{R}=\dfrac{11}{6} \\\\ \sf R=\dfrac{6}{11}$

Therefore, the equivalent resistance of the given combination is 6/11 Ω