Electricity — Question 5

(a) To secure the highest total resistance, the four coils should be connected in series as shown below:

R_eq = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

(b) To secure the lowest total resistance, the four coils should be connected in parallel as shown below:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{24} \\[1em] = \dfrac{6+3+2+1}{24} \\[1em] = \dfrac{12}{24} = \dfrac{1}{2} \Omega$

R_eq = 2Ω

Hence, the lowest total resistance is 2 Ω.