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Electricity · Class 10 · CBSE Science

Three resistors of resistances 2 Omega, 3 Omega, and 6 Omega are connected in parallel. What is the equivalent resistance of the combination?

Practice question from Electricity (Class 10 · CBSE Science), with the correct answer and a step-by-step solution.

medium · MCQ · Q-FBC4B1

Three resistors of resistances 2 Ω2\ \Omega, 3 Ω3\ \Omega, and 6 Ω6\ \Omega are connected in parallel. What is the equivalent resistance of the combination?

  1. A.11 \Omega
  2. B.1 \OmegaCorrect answer
  3. C.6 \Omega
  4. D.0.5 \Omega

Solution

For parallel resistors, 1Rp=1R1+1R2+1R3\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}. Substituting the values: 1Rp=12+13+16=3+2+16=66=1 Ω−1\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1\ \Omega^{-1}. Thus, Rp=1 ΩR_p = 1\ \Omega.

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