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 outlined concepts and calculations from the video transcript.

### Step 1: Understand the given data
- Open circuit potential difference (E) = 10 V
- Load resistances: R1 = 9 Ω and R2 = 4 Ω
- We need to find the heat generated when a load of R = 5 Ω is connected across the battery for a time of T = 1 second.

### Step 2: Use the condition of equal power
When the loads R1 and R2 are connected, the power in both loads is the same. The power (P) across a resistor is given by:
\[ P = V \cdot I \]
Where V is the potential difference across the load and I is the current flowing through it.

Using Ohm's law, we can express V as:
\[ V = I \cdot R \]
Thus, the power can also be expressed as:
\[ P = I^2 \cdot R \]

For the two resistances:
- Power across R1:
\[ P_1 = I_1^2 \cdot R_1 \]
- Power across R2:
\[ P_2 = I_2^2 \cdot R_2 \]

Since \( P_1 = P_2 \), we have:
\[ I_1^2 \cdot R_1 = I_2^2 \cdot R_2 \]

### Step 3: Express currents in terms of E and internal resistance
Using the formula for current through a resistor connected to a battery:
\[ I = \frac{E}{R + r} \]
Where \( r \) is the internal resistance of the battery.

Thus, we can express:
- \( I_1 = \frac{E}{R_1 + r} \)
- \( I_2 = \frac{E}{R_2 + r} \)

Substituting these into the power equality gives:
\[ \left(\frac{E}{R_1 + r}\right)^2 R_1 = \left(\frac{E}{R_2 + r}\right)^2 R_2 \]

### Step 4: Solve for internal resistance (r)
Substituting the values:
\[ \left(\frac{10}{9 + r}\right)^2 \cdot 9 = \left(\frac{10}{4 + r}\right)^2 \cdot 4 \]

Cross-multiplying and simplifying:
\[ 9 \cdot (4 + r)^2 = 4 \cdot (9 + r)^2 \]

Expanding both sides:
\[ 9(16 + 8r + r^2) = 4(81 + 18r + r^2) \]
\[ 144 + 72r + 9r^2 = 324 + 72r + 4r^2 \]

Rearranging gives:
\[ 5r^2 - 180 = 0 \]
\[ r^2 = 36 \]
\[ r = 6 \, \Omega \]

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

### Step 6: Calculate heat generated
Using the formula for heat generated:
\[ H = I^2 \cdot R \cdot T \]
Substituting the values:
\[ H = \left(\frac{10}{11}\right)^2 \cdot 5 \cdot 1 \]
\[ H = \frac{100}{121} \cdot 5 \]
\[ H = \frac{500}{121} \approx 4.13 \, \text{J} \]

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

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