Two batteries of emfs 2 V and 1 V of internal resistances `1Omega and 2Omega` respectively are connected in parallel. The effective emf of the combination is

Options

A. `(3)/(2)V`
B. `(5)/(3)V`
C. `(3)/(5)V`
D. `2 V`

Explanation

To find the effective EMF of two batteries connected in parallel, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- For Battery 1:
- EMF (E1) = 2 V
- Internal Resistance (R1) = 1 Ω
- For Battery 2:
- EMF (E2) = 1 V
- Internal Resistance (R2) = 2 Ω

2. **Use the formula for effective EMF in parallel**:
The formula for the effective EMF (E_effective) when two batteries are connected in parallel is given by:
\[ E_{\text{effective}} = \frac{E_1 R_2 + E_2 R_1}{R_1 + R_2} \]

3. **Substitute the values into the formula**:
\[ E_{\text{effective}} = \frac{(2 \, \text{V} \cdot 2 \, \Omega) + (1 \, \text{V} \cdot 1 \, \Omega)}{1 \, \Omega + 2 \, \Omega} \]

4. **Calculate the numerator**:
- Calculate \(2 \, \text{V} \cdot 2 \, \Omega = 4 \, \text{V} \cdot \Omega\)
- Calculate \(1 \, \text{V} \cdot 1 \, \Omega = 1 \, \text{V} \cdot \Omega\)
- Therefore, the numerator becomes:
\[ 4 \, \text{V} \cdot \Omega + 1 \, \text{V} \cdot \Omega = 5 \, \text{V} \cdot \Omega \]

5. **Calculate the denominator**:
- \(R_1 + R_2 = 1 \, \Omega + 2 \, \Omega = 3 \, \Omega\)

6. **Combine the results**:
\[ E_{\text{effective}} = \frac{5 \, \text{V} \cdot \Omega}{3 \, \Omega} = \frac{5}{3} \, \text{V} \]

7. **Final Result**:
The effective EMF of the combination is:
\[ E_{\text{effective}} = \frac{5}{3} \, \text{V} \]

### Conclusion:
The effective EMF of the combination of the two batteries connected in parallel is \(\frac{5}{3} \, \text{V}\).