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Q100 (NDA-I/2009) Science & Technology › Basic Science (Physics, Chemistry, Biology) › Electricity and magnetism Answer Verified

The resistance of a wire that must be placed parallel with a 12 Q resistance to obtain a combined resistance of 4 Q is

Result
Your answer: —  ·  Correct: C

Explanation

To find the resistance of a wire that must be placed in parallel with a 12 Ω resistor to achieve a combined resistance of 4 Ω, we use the parallel resistance formula: 1/Req = 1/R1 + 1/R2 [c1, t1]. Here, the equivalent resistance (Req) is 4 Ω and one resistor (R1) is 12 Ω [t1, t4]. Substituting these values into the equation gives 1/4 = 1/12 + 1/R2. Rearranging the formula to solve for the unknown resistance, we get 1/R2 = 1/4 - 1/12. By finding a common denominator, 1/R2 = 3/12 - 1/12 = 2/12 [t4]. Simplifying this results in 1/R2 = 1/6, which means R2 = 6 Ω [t4]. Note that the options provided in the question use 'W' (Watts) instead of 'Ω' (Ohms), which is likely a typographical error as the problem describes resistance [c3, t2].

Sources

  1. [1] Science , class X (NCERT 2025 ed.) > Chapter 11: Electricity > Activity 11.6 > p. 186
  2. [2] Science , class X (NCERT 2025 ed.) > Chapter 11: Electricity > What you have learnt > p. 192
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