The series combination of two batteries, both of the same emf 20 V, but different internal resistance of 10 `Omega` and 2 `Omega` , is connected to the parallel combination of two resistors 10 ohm and `ROmega` . 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 and a parallel combination of resistors. Here’s a step-by-step solution:

### Step 1: Understand the Circuit Configuration
We have two batteries in series:
- Battery 1: EMF = 20 V, Internal Resistance (R1) = 10 Ω
- Battery 2: EMF = 20 V, Internal Resistance (R2) = 2 Ω

These batteries are connected to a parallel combination of two resistors:
- Resistor 1 (R1) = 10 Ω
- Resistor 2 (R) = R Ω (unknown)

### Step 2: Analyze the Condition Given
It is given that the voltage difference across the battery with internal resistance 10 Ω (Battery 1) is zero. This means that the terminal voltage of this battery is zero.

### Step 3: Write the Equation for Terminal Voltage
The terminal voltage (V1) for Battery 1 can be expressed as:
\[ V_1 = E_1 - I \cdot R_1 \]
Where:
- \( E_1 = 20 \, V \)
- \( R_1 = 10 \, \Omega \)
- \( I \) is the current flowing through the circuit.

Since \( V_1 = 0 \):
\[ 0 = 20 - I \cdot 10 \]
Rearranging gives:
\[ I \cdot 10 = 20 \]
Thus,
\[ I = \frac{20}{10} = 2 \, A \]

### Step 4: Analyze the Second Battery
Now, we can find the terminal voltage (V2) for Battery 2:
\[ V_2 = E_2 - I \cdot R_2 \]
Where:
- \( E_2 = 20 \, V \)
- \( R_2 = 2 \, \Omega \)

Substituting the values:
\[ V_2 = 20 - 2 \cdot 2 \]
\[ V_2 = 20 - 4 = 16 \, V \]

### Step 5: Set Up the Parallel Circuit
The voltage across both resistors in parallel is the same and equal to the terminal voltage of Battery 2, which is 16 V.

### Step 6: Write the Current Equations for the Parallel Resistors
Let \( I_1 \) be the current through the 10 Ω resistor and \( I_2 \) be the current through the R Ω resistor. The total current \( I \) is the sum of these two:
\[ I = I_1 + I_2 \]
Where:
\[ I_1 = \frac{16}{10} = 1.6 \, A \]
\[ I_2 = \frac{16}{R} \]

### Step 7: Set Up the Equation
Since we know the total current \( I = 2 \, A \):
\[ 2 = 1.6 + \frac{16}{R} \]

### Step 8: Solve for R
Rearranging gives:
\[ \frac{16}{R} = 2 - 1.6 \]
\[ \frac{16}{R} = 0.4 \]

Cross-multiplying gives:
\[ 16 = 0.4R \]
Thus,
\[ R = \frac{16}{0.4} = 40 \, \Omega \]

### Final Answer
The value of \( R \) is **40 Ω**.

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