A galvanometer, together with an unknown resistance in series, is connected across two identical batteries of each `1.5 V`. When the batteries are connected in series,the galvanometer records a current of `1 A`, and when the batteries are connected in parallel, the current is `0.6 A`. In this case, the internal resistance of the battery is

Options

A. `r = (2)/(3) Omega`
B. `r = (2)/(5) Omega`
C. `r = (1)/(3) Omega`
D. `r = (3)/(2) Omega`

Answer

To solve the problem, we need to analyze the two scenarios: when the batteries are connected in series and when they are connected in parallel.

### Step 1: Analyze the Series Connection
When the two batteries are connected in series, the total EMF (E_total) is:
\[ E_{\text{total}} = E + E = 1.5\,V + 1.5\,V = 3\,V \]

Let the internal resistance of each battery be \( r \) and the unknown resistance in series be \( R \). The total internal resistance in series is:
\[ R_{\text{total}} = r + r + R = 2r + R \]

Using Ohm's law, the current \( I \) through the circuit is given by:
\[ I = \frac{E_{\text{total}}}{R_{\text{total}}} = \frac{3}{2r + R} \]
Given that the current \( I = 1\,A \), we can write:
\[ 1 = \frac{3}{2r + R} \]
This simplifies to:
\[ 2r + R = 3 \quad \text{(Equation 1)} \]

### Step 2: Analyze the Parallel Connection
When the two batteries are connected in parallel, the EMF remains the same as one battery:
\[ E_{\text{parallel}} = 1.5\,V \]

The equivalent resistance \( R_{\text{parallel}} \) of the two internal resistances in parallel is:
\[ R_{\text{parallel}} = \frac{r}{2} + R \]

The current \( I \) through the circuit in this case is:
\[ I = \frac{E_{\text{parallel}}}{R_{\text{parallel}}} = \frac{1.5}{\frac{r}{2} + R} \]
Given that the current \( I = 0.6\,A \), we can write:
\[ 0.6 = \frac{1.5}{\frac{r}{2} + R} \]
This simplifies to:
\[ \frac{r}{2} + R = \frac{1.5}{0.6} = 2.5 \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations
Now we have two equations:
1. \( 2r + R = 3 \) (Equation 1)
2. \( \frac{r}{2} + R = 2.5 \) (Equation 2)

From Equation 2, we can express \( R \):
\[ R = 2.5 - \frac{r}{2} \]

Substituting this expression for \( R \) into Equation 1:
\[ 2r + \left(2.5 - \frac{r}{2}\right) = 3 \]
This simplifies to:
\[ 2r + 2.5 - \frac{r}{2} = 3 \]
Combining like terms:
\[ 2r - \frac{r}{2} = 3 - 2.5 \]
\[ \frac{4r}{2} - \frac{r}{2} = 0.5 \]
\[ \frac{3r}{2} = 0.5 \]
Multiplying both sides by \( \frac{2}{3} \):
\[ r = \frac{0.5 \times 2}{3} = \frac{1}{3}\,\Omega \]

### Final Answer
The internal resistance of the battery is:
\[ r = \frac{1}{3}\,\Omega \]