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

Options

A. P1/P2 = 2
B. P1/P2 = 3
C. P1/P2 = 1
D. None

Explanation

To solve the problem, we need to calculate the power used in the `2 Ω` resistor for both circuits separately and then compare them.

### Part (i): Series Circuit with a 6V Battery

1. **Identify the Total Resistance**:
- The resistors are in series: `1 Ω` and `2 Ω`.
- Total resistance \( R_{\text{eq}} = R_1 + R_2 = 1 Ω + 2 Ω = 3 Ω \).

2. **Calculate the Current**:
- Using Ohm's Law, the total current \( I \) in the circuit can be calculated using the formula:
\[ I = \frac{V}{R_{\text{eq}}} \]
- Substituting the values:
\[ I = \frac{6V}{3Ω} = 2 A \]

3. **Calculate the Power in the 2 Ω Resistor**:
- The power \( P \) across the `2 Ω` resistor can be calculated using the formula:
\[ P = I^2 \times R \]
- Substituting the values:
\[ P = (2 A)^2 \times 2 Ω = 4 \times 2 = 8 W \]

### Part (ii): Parallel Circuit with a 4V Battery

1. **Identify the Voltage across the 2 Ω Resistor**:
- In a parallel circuit, the voltage across each resistor is the same as the voltage of the battery.
- Therefore, the voltage across the `2 Ω` resistor is `4V`.

2. **Calculate the Power in the 2 Ω Resistor**:
- The power \( P \) across the `2 Ω` resistor can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
- Substituting the values:
\[ P = \frac{(4V)^2}{2Ω} = \frac{16}{2} = 8 W \]

### Comparison of Power in Both Circuits
- In both cases, the power used in the `2 Ω` resistor is:
- Series Circuit: \( 8 W \)
- Parallel Circuit: \( 8 W \)

Thus, the power used in the `2 Ω` resistor is the same in both circuits.

### Final Answer:
The power used in the `2 Ω` resistor in both circuits is equal, \( 8 W \).

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