A resistor R is conneted to a parallel combination of two identical batteries each with emf E and an internal resistance r. The potential drop across the resistance R is.

Options

A. `(2ER)/(2R+r)`
B. `(ER)/(2R+r)`
C. `(ER)/(2R+r)`
D. `(2ER)/(R+2r)`

Answer

To solve the problem, we need to find the potential drop across the resistor \( R \) when it is connected to a parallel combination of two identical batteries, each with an EMF \( E \) and an internal resistance \( r \).

### Step-by-Step Solution:

1. **Identify the Circuit Configuration**:
We have two identical batteries connected in parallel, each with an EMF \( E \) and an internal resistance \( r \). The external resistor \( R \) is connected across this parallel combination.

2. **Calculate the Equivalent EMF**:
In a parallel circuit, the EMF remains the same as that of one battery. Therefore, the equivalent EMF \( E_{eq} \) of the two batteries is:
\[ E_{eq} = E \]

3. **Calculate the Equivalent Internal Resistance**:
The internal resistances of the two batteries in parallel can be calculated using the formula for resistances in parallel:
\[ \frac{1}{r_{eq}} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r} \]
Thus, the equivalent internal resistance \( r_{eq} \) is:
\[ r_{eq} = \frac{r}{2} \]

4. **Simplify the Circuit**:
Now, we can simplify the circuit to a single battery with EMF \( E \) and an internal resistance \( \frac{r}{2} \) connected to the external resistor \( R \).

5. **Calculate the Total Resistance**:
The total resistance in the circuit is the sum of the equivalent internal resistance and the external resistance:
\[ R_{total} = R + r_{eq} = R + \frac{r}{2} \]

6. **Calculate the Total Current**:
Using Ohm's law, the total current \( I \) flowing through the circuit can be calculated as:
\[ I = \frac{E}{R + \frac{r}{2}} \]

7. **Calculate the Potential Drop Across the Resistor \( R \)**:
The potential drop \( V_R \) across the resistor \( R \) can be calculated using Ohm's law:
\[ V_R = I \cdot R = \left(\frac{E}{R + \frac{r}{2}}\right) \cdot R \]

8. **Final Expression for the Potential Drop**:
Substituting the expression for \( I \) gives:
\[ V_R = \frac{E \cdot R}{R + \frac{r}{2}} \]

### Final Answer:
The potential drop across the resistance \( R \) is:
\[ V_R = \frac{E \cdot R}{R + \frac{r}{2}} \]