Compare the power used in the \(2 \Omega\) resistor in each of the following circuits: (i) a \(6 \mathrm{~V}\) battery in series with \(1 \Omega\) and \(2 \Omega\) resistors, and (ii) a \(4 \mathrm{~V}\) battery in parallel with \(12 \Omega\) and \(2 \Omega\) resistors.

Ohm's Law is a fundamental principle in electrical circuits. It states that the current through a conductor between two points is directly proportional to the voltage across the two points. The formula is:
\[ V = I \times R \ \text{where} \ V \text{ is the voltage} \ I \text{ is the current} \ R \text{ is the resistance} \]
In the given exercise, Ohm's Law is used to calculate the current in the series circuit. By rearranging the formula
\[ I = \frac{V}{R} \ \text{we get} \ I_{\text{series}} = \frac{6 \text{ V}}{3 \text{ Ω}} = 2 \text{ A} \]
This shows that a 6V battery in a series circuit with 3 ohms of resistance will produce a current of 2 amperes.

In a series circuit, all components are connected end-to-end to form one path for current to flow. The total resistance in a series circuit is the sum of the individual resistances.
In this exercise, the resistors of 1 Ω and 2 Ω are connected in series. The total resistance is
\[ R_{\text{total}} = R_1 + R_2 = 1 \text{ Ω} + 2 \text{ Ω} = 3 \text{ Ω} \]
The current through each resistor is the same. Using Ohm's Law, the current can then be calculated using the total resistance and the supplied voltage. Here, the current in the series circuit is
\[ I_{\text{series}} = \frac{6 \text{ V}}{3 \text{ Ω}} = 2 \text{ A} \]
Thus, every resistor in this series circuit has 2 amperes of current flowing through it.

In parallel circuits, all components are connected across each other, creating multiple paths for the current. The voltage across each resistor in a parallel circuit is the same.
For the exercise, the 12 Ω and 2 Ω resistors are connected in parallel with a 4V battery. Thus, the voltage across every resistor is 4V.
The power consumed by the 2 Ω resistor in the parallel circuit can be calculated using the power formula
\[ P = \frac{V^2}{R} \ \text{which gives} \ P_{\text{parallel}} = \frac{(4 \text{ V})^2}{2 \text{ Ω}} = 8 \text{ W} \]
Each resistor receives the full voltage from the battery, leading to unique currents through each path according to their resistance.

Electrical power is the rate at which electrical energy is converted to another form of energy (like heat). It can be calculated using different forms of the basic formula
\[ P = V \times I \ \text{or} \ P = I^2 \times R \ \text{or} \ P = \frac{V^2}{R} \]In the series circuit, the power used by the 2 Ω resistor is calculated with
\[ P_{\text{series}} = I^2 \times R = (2 \text{ A})^2 \times 2 \text{ Ω} = 8 \text{ W} \]
In the parallel circuit, using the voltage and resistance, we find the power:
\[ P_{\text{parallel}} = \frac{V^2}{R} = \frac{(4 \text{ V})^2}{2 \text{ Ω}} = 8 \text{ W} \]
This shows that despite the difference in circuit configuration, both the series and parallel configurations result in the 2 Ω resistor using 8 watts of power.