Supply voltage: \(V= 220 \ V\)
Resistance of a single coil: \(R= 24\ Ω\)
Scenario (i): Individual coil usage
Using Ohm's law: \(V= I_1R_1\)
Where \(I_1\) is the current through the coil.
\(I_1 = \frac {V}{R_1}\) is calculated as \(\frac {220}{24}\), resulting in \(I_1= 9.166 \ A\).
Thus, 9.16 A flows through the coil when used individually.
Scenario (ii): Coils in series
Total resistance: \(R_2 = 24 Ω + 24 Ω = 48 Ω\)
From Ohm's law: \(V = I_2R_2\)
Where \(I_2\) is the current in the series circuit.
\(I_2 = \frac {V}{R_2}\) is calculated as \(\frac {220}{48}\), resulting in \(I_2 = 4.58\ A\).
Therefore, 4.58 A flows through the circuit when coils are connected in series.
Scenario (iii): Coils in parallel
Total resistance: \(R_3\) = \(\frac {1}{\frac {1}{24}+\frac {1}{24}}\) = \(\frac {24}{2}\) = \(12\ Ω\)
According to Ohm's law: \(V= I_3R_3\)
Where \(I_3\) is the current flowing through the circuit.
\(I_3 = \frac {V}{R_3}\) is calculated as \(\frac {220}{12}\), resulting in \(I_3= 18.33\ A\).
Hence, 18.33 A flows through the circuit when coils are connected in parallel.