Series And Parallel Combination
Two resistors of 6 (\Omega) and 3 (\Omega) are joined in parallel, and this combination is connected in series with a 4 (\Omega) resistor. What is the net resistance of the arrangement?
Select the correct option:
Solution
6 \(\Omega\)
Resistor networks are reduced by treating parallel branches, where the reciprocal of the equivalent resistance equals the sum of reciprocals, and series elements, where resistances simply add. First the parallel pair is combined: (\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6}), giving (R_p = 2;\Omega). This equivalent then sits in series with the 4 (\Omega) resistor, so the total is (R = R_p + 4 = 2 + 4 = 6;\Omega). The value 13 (\Omega) is wrong because it adds all three resistors as if entirely in series. The value 2 (\Omega) ignores the series resistor and reports only the parallel section. The value 9 (\Omega) wrongly takes the parallel pair as 5 (\Omega). This applies the NCERT rules for combining resistors in mixed networks. A plausibility check confirms the result, since the parallel equivalent must be smaller than the smallest branch (less than 3 (\Omega)), and adding 4 (\Omega) in series sensibly raises the total to 6 (\Omega). Conceptually, the parallel section offers the charge two alternative routes and so lowers the opposition, whereas the series resistor lies on the single common path that every electron must traverse, which is why the two combination rules differ in form yet are applied sequentially to collapse the whole network into one equivalent value.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- physics
- Chapter
- current electricity
- Topic
- series and parallel combination
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
6 \(\Omega\)
Resistor networks are reduced by treating parallel branches, where the reciprocal of the equivalent resistance equals the sum of reciprocals, and series elements, where resistances simply add. First the parallel pair is combined: (\frac{1}{R_p} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6}), giving (R_p = 2;\Omega). This equivalent then sits in series with the 4 (\Omega) resistor, so the total is (R = R_p + 4 = 2 + 4 = 6;\Omega). The value 13 (\Omega) is wrong because it adds all three resistors as if entirely in series. The value 2 (\Omega) ignores the series resistor and reports only the parallel section. The value 9 (\Omega) wrongly takes the parallel pair as 5 (\Omega). This applies the NCERT rules for combining resistors in mixed networks. A plausibility check confirms the result, since the parallel equivalent must be smaller than the smallest branch (less than 3 (\Omega)), and adding 4 (\Omega) in series sensibly raises the total to 6 (\Omega). Conceptually, the parallel section offers the charge two alternative routes and so lowers the opposition, whereas the series resistor lies on the single common path that every electron must traverse, which is why the two combination rules differ in form yet are applied sequentially to collapse the whole network into one equivalent value.
This medium difficulty physics question is from the chapter current electricity, covering the topic of series and parallel combination. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse current electricity questions on RankGuru.