A resistance `R_(1)` is connected to a source of constant voltage . On connecting a resistance `R_(2)` in series with `R_(1)`

Options

A. The total thermal power dissipated increases
B. Thermal power dissipated by `R_(1)` decreases
C. Thermal power dissipated by `R_(1)` increases
D. Total power dissipiated decreases

Answer

To solve the problem step-by-step, let's analyze the situation where a resistance \( R_1 \) is connected to a constant voltage source, and then a resistance \( R_2 \) is added in series with \( R_1 \).

### Step 1: Understand the Circuit with Only \( R_1 \)
When only \( R_1 \) is connected to the voltage source with an EMF \( E \):
- The current \( I \) flowing through the circuit is given by Ohm's Law:
\[ I = \frac{E}{R_1} \]
- The power \( P \) dissipated in \( R_1 \) is given by the formula:
\[ P_{R_1} = I^2 R_1 = \left(\frac{E}{R_1}\right)^2 R_1 = \frac{E^2}{R_1} \]

### Step 2: Analyze the Circuit with Both \( R_1 \) and \( R_2 \)
Now, when \( R_2 \) is connected in series with \( R_1 \):
- The total resistance in the circuit becomes \( R_1 + R_2 \).
- The new current \( I' \) flowing through the circuit is:
\[ I' = \frac{E}{R_1 + R_2} \]
- The total power \( P_{total} \) dissipated in the circuit is:
\[ P_{total} = I'^2 (R_1 + R_2) = \left(\frac{E}{R_1 + R_2}\right)^2 (R_1 + R_2) = \frac{E^2}{R_1 + R_2} \]

### Step 3: Calculate the Power Across \( R_1 \) with Both Resistors
The power dissipated across \( R_1 \) when both resistors are connected is:
- Using the current \( I' \):
\[ P_{R_1} = I'^2 R_1 = \left(\frac{E}{R_1 + R_2}\right)^2 R_1 = \frac{E^2 R_1}{(R_1 + R_2)^2} \]

### Step 4: Compare the Powers
1. **Total Power Comparison**:
- When only \( R_1 \) is connected: \( P_{total} = \frac{E^2}{R_1} \)
- When both \( R_1 \) and \( R_2 \) are connected: \( P_{total} = \frac{E^2}{R_1 + R_2} \)
- Since \( R_1 + R_2 > R_1 \), it follows that:
\[ P_{total} < \frac{E^2}{R_1} \quad \text{(Total power decreases)} \]

2. **Power Across \( R_1 \) Comparison**:
- When only \( R_1 \) is connected: \( P_{R_1} = \frac{E^2}{R_1} \)
- When both resistors are connected: \( P_{R_1} = \frac{E^2 R_1}{(R_1 + R_2)^2} \)
- We cannot definitively compare \( P_{R_1} \) in both cases without knowing the relationship between \( R_1 \) and \( R_2 \).

### Conclusion
- The total thermal power dissipated in the circuit decreases when \( R_2 \) is added in series.
- We cannot determine whether the power across \( R_1 \) increases or decreases without additional information about the values of \( R_1 \) and \( R_2 \).

### Final Answer
- The total thermal power decreases when \( R_2 \) is added in series with \( R_1 \).
- We cannot comment on the power dissipated along \( R_1 \).