Resistors $R_{1} = 3 Ω$ , $R_{2} = 6 Ω$ , and $R_{3} = 6 Ω$ are connected such that $R_{1}$ is in parallel with the series combination of $R_{2}$ and $R_{3}$ . Find the equivalent resistance of the whole network.

hard 5 marks Class 10 · Science · Electricity

Answer

R_{e q} = 2.4 Ω

The correct answer is $R_{e q} = 2.4 Ω$ — Invert to obtain $R_{e q} = \frac{12}{5} = 2.4 Ω$ .

Step-by-step solution

First find the series resistance of $R_{2}$ and $R_{3}$ : $R_{ser i es} = R_{2} + R_{3} = 6 + 6 = 12 Ω$ .
Now $R_{1}$ is in parallel with $R_{ser i es}$ . Use the parallel formula $\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{ser i es}}$ .
Substitute $R_{1} = 3 Ω$ and $R_{ser i es} = 12 Ω$ : $\frac{1}{R _{e q}} = \frac{1}{3} + \frac{1}{12} = \frac{4}{12} + \frac{1}{12} = \frac{5}{12}$ .
Invert to obtain $R_{e q} = \frac{12}{5} = 2.4 Ω$ .

Common mistake: Adding all three resistances directly or using the parallel formula with wrong pairings.

Concept tested: Series‑parallel combination of resistors