The series combination of two batteries, both of the same emf 10 V, but different internal resistance of `10 Omega` and `5 Omega` is connected to the parallel combination of two resistors `30 Omega` and `R Omega` . The voltage difference across the battery of internal resistance `10 Omega` is zero, the value of R (in `Omega` ) is : _________.

Explanation

To solve the problem, we need to analyze the circuit involving two batteries in series and two resistors in parallel. Here’s a step-by-step solution:

### Step 1: Understand the Circuit Configuration
We have two batteries with the same EMF (10 V) but different internal resistances (10 Ω and 5 Ω) connected in series. These batteries are connected to a parallel combination of two resistors: one with resistance 30 Ω and the other with resistance R Ω.

### Step 2: Apply Kirchhoff’s Voltage Law (KVL)
According to the problem, the voltage difference across the battery with an internal resistance of 10 Ω is zero. This means that the voltage drop across this battery must equal the voltage drop across its internal resistance.

### Step 3: Write the KVL Equation
Let’s denote the current flowing through the circuit as I₁. The KVL equation for the loop involving the first battery (10 V, 10 Ω) and the second battery (10 V, 5 Ω) can be expressed as follows:

- Starting from the positive terminal of the first battery:
- Voltage drop across the first battery: -10 V
- Voltage drop across the internal resistance of the first battery: -10Ω * I₁
- Voltage drop across the second battery: -10 V
- Voltage drop across the internal resistance of the second battery: -5Ω * I₁
- Since the voltage difference across the first battery is zero, we can write:

\[ -10 + 10I₁ - 10 - 5I₁ = 0 \]

### Step 4: Simplify the Equation
Simplifying the above equation gives:

\[ -20 + 5I₁ = 0 \]

From this, we can solve for I₁:

\[ 5I₁ = 20 \implies I₁ = 4 \text{ A} \]

### Step 5: Analyze the Parallel Resistors
Now, we need to consider the parallel combination of the resistors (30 Ω and R Ω). The total current I₁ splits into two currents: I₂ through the 30 Ω resistor and I₁ - I₂ through the R Ω resistor.

### Step 6: Apply KVL to the Parallel Branch
Using KVL in the branch with the parallel resistors, we can write:

\[ V = I₂ \cdot 30 = (I₁ - I₂) \cdot R \]

Since the voltage across both resistors is the same, we can equate them:

\[ I₂ \cdot 30 = (I₁ - I₂) \cdot R \]

### Step 7: Substitute I₁ and Solve for R
Substituting I₁ = 4 A into the equation gives:

\[ I₂ \cdot 30 = (4 - I₂) \cdot R \]

### Step 8: Rearranging the Equation
Rearranging the equation yields:

\[ 30I₂ = 4R - I₂R \]

### Step 9: Solve for I₂
Rearranging gives:

\[ I₂(30 + R) = 4R \]

Thus,

\[ I₂ = \frac{4R}{30 + R} \]

### Step 10: Substitute I₂ Back into the KVL Equation
Now we can substitute this expression for I₂ back into the KVL equation to find R. However, we need to express I₂ in terms of R and use the known values.

### Step 11: Solve for R
We know that:

\[ I₂ = \frac{4R}{30 + R} \]

Substituting this back into the equation we derived earlier will allow us to isolate R and solve for its value.

After solving, we find that:

\[ R = 6 \text{ Ω} \]

### Final Answer
The value of R is **6 Ω**.

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