Two cells having an internal resistance of `0.2 Omega` and `0.4 Omega` are connected in parallel (same polarity) the voltage across the battery is 1.5 V. If the emf of one cell is 1.2 V, then the emf of second cell is

Answer

To solve the problem, we need to find the EMF of the second cell (E2) when two cells with internal resistances of 0.2 Ω and 0.4 Ω are connected in parallel, and the voltage across the battery is 1.5 V. One cell has an EMF of 1.2 V.

### Step-by-Step Solution:

1. **Understanding the Circuit:**
- We have two cells connected in parallel with the same polarity.
- The voltage across the combination is given as 1.5 V.
- The EMF of the first cell (E1) is 1.2 V, and its internal resistance (R1) is 0.2 Ω.
- The internal resistance of the second cell (R2) is 0.4 Ω, and we need to find its EMF (E2).

2. **Applying Kirchhoff's Voltage Law:**
- For the first cell (E1), the voltage across it can be expressed as:
\[ V = E1 - I \cdot R1 \]
- For the second cell (E2), the voltage across it can be expressed as:
\[ V = E2 + I \cdot R2 \]
- Since both cells are in parallel, the voltage across both cells is the same (1.5 V).

3. **Setting Up the Equations:**
- For the first cell:
\[ 1.5 = 1.2 - I \cdot 0.2 \quad \text{(1)} \]
- For the second cell:
\[ 1.5 = E2 + I \cdot 0.4 \quad \text{(2)} \]

4. **Solving Equation (1) for Current (I):**
- Rearranging equation (1):
\[ I \cdot 0.2 = 1.2 - 1.5 \]
\[ I \cdot 0.2 = -0.3 \]
\[ I = \frac{-0.3}{0.2} = -1.5 \, \text{A} \]

5. **Substituting I into Equation (2):**
- Substitute I into equation (2):
\[ 1.5 = E2 + (-1.5) \cdot 0.4 \]
\[ 1.5 = E2 - 0.6 \]
\[ E2 = 1.5 + 0.6 = 2.1 \, \text{V} \]

### Final Answer:
The EMF of the second cell (E2) is **2.1 V**.

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