2024 Current ElectricitySeries-Parallel Combination of Resistors Medium Question 1 of 2 in Series-Parallel Combination of Resistors

A wire of length '$l$' and resistance 100 $\Omega$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is;

A 55 $\Omega$
B 60 $\Omega$
C 26 $\Omega$
D 52 $\Omega$

The answer is D.

Solution

Resistance of each part $R = \frac{100}{10} = 10\ \Omega$ Series connection of 5 parts: $R_s = nR = 5 \times 10 = 50\ \Omega$ Parallel connection of 5 parts: $R_p = \frac{R}{n} = \frac{10}{5} = 2\ \Omega$ $R_{equivalent} = R_s + R_p = 50 + 2 = 52\ \Omega$

Concept Combining series and parallel segments of a divided wire

Series-Parallel Combination of Resistors in past papers

2 questions from this chapter have appeared across 2 exam years.