Question

You are given three resistances of 1, 2 and 3 ohms. Shows by diagrams, how with the help of these resistances you can get:

(i) 6 Ω
(ii) $\frac{6}{11}$ Ω
(iii) 1.5 Ω

Answer 1

(i) To get a value of 6 Ω, all the resistors are to be connected in series as shown below:

1 Ώ 2 Ώ 3 Ώ


Net resistance, R = R₁ + R₂ + R₃
R = 1 Ώ + 2 Ώ + 3 Ώ = 6 Ώ

(ii) To get a value of 6 / 11 Ω, all the resistors are to be connected in parallel as shown below:

$$\begin{gathered} \mathrm{When}\mathrm{three}\mathrm{resistance}\mathrm{is}\mathrm{connected}\mathrm{in}\mathrm{parallel} \\ \mathrm{their}\mathrm{resultant}\mathrm{resistance}\mathrm{is}\mathrm{given}\mathrm{by} \\ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{{\mathrm{R}}_3} \\ {R}_{1=}3\mathrm{\Omega } \\ {R}_2=2\mathrm{\Omega } \\ {R}_3=1\mathrm{\Omega } \\ \frac{1}{R}=\frac{1}{3}+\frac{1}{2}+\frac{1}{1} \\ \frac{1}{R}=\frac{2+3+6}{6} \\ \frac{1}{R}=\frac{11}{6} \\ R=\frac{6}{11}\mathrm{\Omega } \end{gathered}$$

(iii) To get a value of 1.5 Ω, the 1 Ω and 2 Ω resistors should be connected in series and this arrangement should be connected in parallel with the 3 Ω resistor as shown below:

The resistors of resistance 1 Ώ and 2 Ώ are in series. Therefore, their net resistance is:
R = R₁ + R₂
R = 1 Ώ + 2 Ώ
R = 3 Ώ
This 3 Ώ is connected in parallel with another 3 Ώ resistance.
Therefore, net resistance will be:
$$\begin{gathered} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R}=\frac{1}{3}+\frac{1}{3} \\ \frac{1}{R}=\frac{2}{3} \\ R=\frac{3}{2}=1.5\mathrm{\Omega } \end{gathered}$$