Two batteries with emf 12V and 13V are connected in parallel across a load resistor of `10Omega`. The internal resistances of the two batteries are `1 Omega and 2 Omega` respectively. The voltage across the load lies between

Options

A. 11.4V and 11.5V
B. 11.7 V and 11.8V
C. 11.6V and 11.7V
D. 11.5 V and 11.6V

Explanation

To solve the problem of finding the voltage across the load resistor when two batteries are connected in parallel, we will follow these steps:

### Step 1: Identify the given values
- EMF of Battery 1 (E1) = 12 V
- Internal resistance of Battery 1 (R1) = 1 Ω
- EMF of Battery 2 (E2) = 13 V
- Internal resistance of Battery 2 (R2) = 2 Ω
- Load resistance (R_load) = 10 Ω

### Step 2: Calculate the equivalent EMF (E_eq)
The formula for the equivalent EMF when batteries are connected in parallel is given by:
\[ E_{eq} = \frac{E_1 R_2 + E_2 R_1}{R_1 + R_2} \]
Substituting the values:
\[ E_{eq} = \frac{12 \times 2 + 13 \times 1}{1 + 2} = \frac{24 + 13}{3} = \frac{37}{3} \text{ V} \approx 12.33 \text{ V} \]

### Step 3: Calculate the equivalent internal resistance (R_eq)
The equivalent internal resistance for two resistors in parallel is given by:
\[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \]
Substituting the values:
\[ R_{eq} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \text{ Ω} \approx 0.67 \text{ Ω} \]

### Step 4: Calculate the total resistance in the circuit
The total resistance in the circuit (R_total) is the sum of the equivalent internal resistance and the load resistance:
\[ R_{total} = R_{eq} + R_{load} = \frac{2}{3} + 10 = \frac{2}{3} + \frac{30}{3} = \frac{32}{3} \text{ Ω} \approx 10.67 \text{ Ω} \]

### Step 5: Calculate the total current (I)
Using Ohm’s law, the total current flowing through the circuit can be calculated as:
\[ I = \frac{E_{eq}}{R_{total}} = \frac{\frac{37}{3}}{\frac{32}{3}} = \frac{37}{32} \text{ A} \approx 1.15625 \text{ A} \]

### Step 6: Calculate the voltage across the load resistor (V_load)
The voltage across the load resistor can be calculated using Ohm’s law:
\[ V_{load} = I \times R_{load} = \frac{37}{32} \times 10 = \frac{370}{32} \text{ V} \approx 11.56 \text{ V} \]

### Conclusion
The voltage across the load lies between 11.5 V and 11.6 V.