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 find the potential drop across the resistor \( R \) connected to a parallel combination of two identical batteries, we can follow these steps:

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

### Step 2: Calculate the Equivalent EMF of the Batteries
For two identical batteries in parallel, the equivalent EMF \( E_{eq} \) remains the same as the individual EMF:
\[ E_{eq} = E \]

### Step 3: Calculate the Equivalent Internal Resistance
The internal resistances of the two batteries in parallel combine as follows:
\[ r_{eq} = \frac{r}{2} \]

### Step 4: Set Up the Circuit with Equivalent Values
Now, we can represent the circuit with the equivalent EMF and equivalent internal resistance:
- EMF: \( E_{eq} = E \)
- Internal Resistance: \( r_{eq} = \frac{r}{2} \)

The circuit now looks like this:
- A battery with EMF \( E \) and internal resistance \( \frac{r}{2} \) in series with the resistor \( R \).

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

### Step 6: Calculate the Current \( I \) in the Circuit
Using Ohm's law, the current \( I \) flowing through the circuit can be calculated as:
\[ I = \frac{E_{eq}}{R_{total}} = \frac{E}{R + \frac{r}{2}} \]

### Step 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:
\[ V_R = I \cdot R = \left(\frac{E}{R + \frac{r}{2}}\right) \cdot R \]
Substituting the value of \( I \):
\[ V_R = \frac{E \cdot R}{R + \frac{r}{2}} \]

### Step 8: Simplify the Expression
To simplify the expression, we can multiply the numerator and denominator by 2:
\[ V_R = \frac{2ER}{2R + r} \]

### Final Result
The potential drop across the resistor \( R \) is:
\[ V_R = \frac{2ER}{2R + r} \]

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