Electricity — Question 13

Case 1 — When coils are used separately

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{220}{24} \\[1em] \Rightarrow \text{I} = 9.166 \text{ A} \approx 9.2 \text{ A}$

Hence, 9.2 A of current flows through each resistor when they are used separately.

Case 2 — When coils are connected in series

R_s = 24 Ω + 24 Ω = 48 Ω

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{220}{48} \\[1em] \Rightarrow \text{I} = 4.58 \text{ A} \approx 4.6 \text{ A}$

Therefore, the current is 4.6 A when the two coils are used in series.

Case 3 — When coils are connected in parallel

$\dfrac{1}{\text{R}_p} = \dfrac{1}{24} + \dfrac{1}{24} = \dfrac{2}{24} = \dfrac{1}{12}$

∴ R_p = 12 Ω

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{220}{12} \\[1em] \Rightarrow \text{I} = 18.3 \text{ A}$

Therefore, the current is 18.3 A when the two coils are used in parallel.