Question

The figure given below shows three resistors?



Their combined resistance is :

(a) $1\frac{5}{7}$ Ω
(b) 14 Ω
(c) $6\frac{2}{3}$ Ω
(d) $7\frac{1}{2}$ Ω

Answer 1

(d) 7$\frac{1}{2}$$\mathrm{\Omega }$
The resistors of 6 Ώ and 2 Ώ are connected in parallel.
$$\begin{gathered} \therefore \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \mathrm{Here}, \\ {R}_1=6\mathrm{\Omega } \\ {R}_2=2\mathrm{\Omega } \\ \frac{1}{R}=\frac{1}{6}+\frac{1}{2} \\ \frac{1}{R}=\frac{4}{6} \\ R=\frac{6}{4} \end{gathered}$$

This arrangement is further connected in series with the 6 Ώ resistor.
∴ Net resistance =$\frac{6}{4}+6$ =$7\frac{1}{2}$ Ω