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Question

Three wires of resistance 3 ohm, 6 ohm and 9 ohm are connected in a parallel. What will be the total resistance of the circuit?

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is

$\frac{18}{11}$ohms 

Parallel Resistance Calculation for Three Wires

To find the total resistance ($R_{total}$) of resistors connected in parallel, we use the formula:

$ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $

Given the resistances:

  • $R_1 = 3 \Omega$
  • $R_2 = 6 \Omega$
  • $R_3 = 9 \Omega$

Substitute these values into the formula:

$ \frac{1}{R_{total}} = \frac{1}{3} + \frac{1}{6} + \frac{1}{9} $

To add these fractions, find a common denominator, which is 18:

$ \frac{1}{R_{total}} = \frac{1 \times 6}{3 \times 6} + \frac{1 \times 3}{6 \times 3} + \frac{1 \times 2}{9 \times 2} $ $ \frac{1}{R_{total}} = \frac{6}{18} + \frac{3}{18} + \frac{2}{18} $

Add the numerators:

$ \frac{1}{R_{total}} = \frac{6 + 3 + 2}{18} $ $ \frac{1}{R_{total}} = \frac{11}{18} $

Finally, invert the result to find the total resistance ($R_{total}$):

$ R_{total} = \frac{18}{11} \Omega $

Therefore, the total resistance of the circuit is $\frac{18}{11}$ ohms.

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