A resistor `R_(1)` dissipates the power `P` when connected to a certain generator. If the resistor `R_(2)` is put in series with `R_(1)`, the power dissipated by `R_(1)`

Options

A. decreases
B. increases
C. remains the same
D. any of the above depending upon the relative values of `R_(1)` and `R_(2)`

Answer

To solve the problem, we need to analyze the power dissipated by resistor \( R_1 \) when it is connected to a generator and then when a second resistor \( R_2 \) is added in series with it.

### Step-by-Step Solution:

1. **Initial Condition with Resistor \( R_1 \)**:
- When only resistor \( R_1 \) is connected to the generator, the power \( P \) dissipated by \( R_1 \) can be expressed using the formula:
\[ P = I^2 R_1 \]
- Here, \( I \) is the current flowing through \( R_1 \).

2. **Current Calculation**:
- The current \( I \) can be calculated using Ohm's law, where \( V \) is the voltage provided by the generator:
\[ I = \frac{V}{R_1} \]
- Substituting this into the power formula gives:
\[ P = \left(\frac{V}{R_1}\right)^2 R_1 = \frac{V^2}{R_1} \]

3. **Adding Resistor \( R_2 \) in Series**:
- When resistor \( R_2 \) is added in series with \( R_1 \), the total resistance \( R_{total} \) becomes:
\[ R_{total} = R_1 + R_2 \]
- The new current \( I' \) flowing through the circuit is now:
\[ I' = \frac{V}{R_1 + R_2} \]

4. **Power Dissipated by \( R_1 \) with \( R_2 \) in Series**:
- The power \( P' \) dissipated by \( R_1 \) when \( R_2 \) is in series can be expressed as:
\[ P' = (I')^2 R_1 \]
- Substituting \( I' \) into the power formula gives:
\[ P' = \left(\frac{V}{R_1 + R_2}\right)^2 R_1 = \frac{V^2 R_1}{(R_1 + R_2)^2} \]

5. **Comparison of Powers**:
- To compare \( P' \) and \( P \):
\[ P' = \frac{V^2 R_1}{(R_1 + R_2)^2} \quad \text{and} \quad P = \frac{V^2}{R_1} \]
- Since \( R_1 + R_2 > R_1 \), it follows that \( (R_1 + R_2)^2 > R_1^2 \). Therefore:
\[ P' < P \]
- This means the power dissipated by \( R_1 \) decreases when \( R_2 \) is added in series.

### Conclusion:
The power dissipated by resistor \( R_1 \) decreases when resistor \( R_2 \) is added in series.

### Final Answer:
**The power dissipated by \( R_1 \) decreases.**