Question

What is (a) highest, and (b) lowest, resistance which can be obtained by combining four resistors having the following resistances?

4 Ω, 8 Ω, 12 Ω, 24 Ω

Answer 1

(a) To get the highest resistance, all the resistors must be connected in series.
Resistance in a series arrangement is given by R = R₁+ R₂ + R₃ + R₄
$$\begin{gathered} R=4\mathrm{\Omega }+8\mathrm{\Omega }+12\mathrm{\Omega }+24\mathrm{\Omega } \\ R=48\mathrm{\Omega } \end{gathered}$$
Therefore, the highest resistance is 48 Ω.
(b) To get the lowest resistance, all the resistors must be connected in parallel.
Resistance in a parallel arrangement is given by:
$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$

Here R₁= 4 $\mathrm{\Omega }$,
R₂ = 8 $\mathrm{\Omega }$,
R₃= 12 $\mathrm{\Omega }$,
R₄ = 24 $\mathrm{\Omega }$

$$\begin{gathered} \frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24} \\ \frac{1}{R}=\frac{6+3+2+1}{24} \\ \frac{1}{R}=\frac{12}{24} \\ R=2\mathrm{\Omega } \end{gathered}$$
Therefore, the lowest resistance of the arrangement is 2 Ω.