A battery has an open circuit potential difference 10 V between the terminals. When loads `9Omega and 4 Omega` are connected one by one across the battery , the power in the load resistance is the same. The amount of heat approximately generated in one second in the load when a load of `5Omega` Is connected across the battery will be

Options

A. `10.4J`
B. 9 J
C. `6.8 J`
D. `4.13J`

Answer

To solve the problem step by step, we will follow the logical sequence of calculations as outlined in the video transcript.

### Step 1: Identify Given Values
- Open circuit potential difference (EMF, E) = 10 V
- Load resistances: R1 = 9 Ω, R2 = 4 Ω
- Load resistance for which we need to find heat (RL) = 5 Ω

### Step 2: Key Concept for Heat Generation
The heat generated in a resistor can be calculated using the formula:
\[ H = V \cdot I \cdot t \]
Since \( V = I \cdot R \), we can also express heat as:
\[ H = I^2 \cdot R \cdot t \]
Where:
- \( H \) = heat generated
- \( I \) = current through the load
- \( R \) = resistance of the load
- \( t \) = time (in seconds)

### Step 3: Find Internal Resistance
To find the internal resistance of the battery, we need to use the condition that the power in the load resistances (9 Ω and 4 Ω) is the same when connected one by one.

Using the power formula:
\[ P = I^2 \cdot R \]
For the two loads:
\[ P_1 = I_1^2 \cdot R_1 \]
\[ P_2 = I_2^2 \cdot R_2 \]
Since \( P_1 = P_2 \):
\[ I_1^2 \cdot 9 = I_2^2 \cdot 4 \]

### Step 4: Express Currents in Terms of EMF and Resistances
Using Ohm's law, we can express the currents as:
\[ I_1 = \frac{E}{R_1 + r} = \frac{10}{9 + r} \]
\[ I_2 = \frac{E}{R_2 + r} = \frac{10}{4 + r} \]

### Step 5: Set Up the Equation
Substituting \( I_1 \) and \( I_2 \) into the power equation:
\[ \left(\frac{10}{9 + r}\right)^2 \cdot 9 = \left(\frac{10}{4 + r}\right)^2 \cdot 4 \]

### Step 6: Cross Multiply and Simplify
Cross-multiplying gives:
\[ 100 \cdot 9 \cdot (4 + r)^2 = 100 \cdot 4 \cdot (9 + r)^2 \]
This simplifies to:
\[ 9(4 + r)^2 = 4(9 + r)^2 \]

### Step 7: Expand and Solve for r
Expanding both sides:
\[ 9(16 + 8r + r^2) = 4(81 + 18r + r^2) \]
This leads to:
\[ 144 + 72r + 9r^2 = 324 + 72r + 4r^2 \]
Simplifying gives:
\[ 5r^2 - 180 = 0 \]
Thus:
\[ r^2 = 36 \implies r = 6 \, \Omega \]

### Step 8: Calculate Current for 5 Ω Load
Now, we can find the current when a 5 Ω load is connected:
\[ I = \frac{E}{R_L + r} = \frac{10}{5 + 6} = \frac{10}{11} \, \text{A} \]

### Step 9: Calculate Heat Generated
Using the heat formula:
\[ H = I^2 \cdot R_L \cdot t = \left(\frac{10}{11}\right)^2 \cdot 5 \cdot 1 \]
Calculating this gives:
\[ H = \frac{100}{121} \cdot 5 \approx 4.13 \, \text{J} \]

### Final Answer
The amount of heat approximately generated in one second in the load when a load of 5 Ω is connected across the battery is **approximately 4.13 J**.