An unknown resistor is connected between the terminals of a 3.00 V battery. Energy is dissipated in the resistor at the rate of 0.707 W. The same resistor is then connected between the terminals of a 12.0 V battery. At what rate is energy now dissipated?

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the relationship between power, voltage, and resistance.
The power dissipated in a resistor can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
where \( P \) is the power (in watts), \( V \) is the voltage (in volts), and \( R \) is the resistance (in ohms).

### Step 2: Calculate the resistance using the first battery.
Given:
- Voltage \( V_1 = 3.00 \, \text{V} \)
- Power \( P_1 = 0.707 \, \text{W} \)

We can rearrange the power formula to find the resistance \( R \):
\[ R = \frac{V^2}{P} \]
Substituting the known values:
\[ R = \frac{(3.00)^2}{0.707} = \frac{9.00}{0.707} \approx 12.72 \, \Omega \]

### Step 3: Calculate the power dissipated with the second battery.
Now we will use the resistance we just calculated to find the power dissipated when the resistor is connected to a 12.0 V battery.

Given:
- Voltage \( V_2 = 12.0 \, \text{V} \)

Using the power formula again:
\[ P_2 = \frac{V_2^2}{R} \]
Substituting the known values:
\[ P_2 = \frac{(12.0)^2}{12.72} = \frac{144}{12.72} \approx 11.3 \, \text{W} \]

### Final Answer:
The rate at which energy is dissipated when the resistor is connected to the 12.0 V battery is approximately \( 11.3 \, \text{W} \).

---