When two identical cells are connected either in series or in parallel across a 4 ohm resistor, they send the same current through it. The internal resistance of the cell in ohm is

Options

A. `4Omega`
B. `2Omega`
C. `1Omega`
D. `7Omega`

Explanation

To solve the problem of finding the internal resistance of the cell when two identical cells are connected either in series or in parallel across a 4-ohm resistor, we can follow these steps:

### Step 1: Understand the Configuration
We have two identical cells, each with an electromotive force (EMF) \( E \) and an internal resistance \( r \). We will analyze the two configurations: series and parallel.

### Step 2: Series Connection
In a series connection, the total voltage across the resistor is the sum of the voltages of the two cells:
\[ V_{\text{total}} = E + E = 2E \]
The total resistance in the circuit is the sum of the internal resistances and the external resistor:
\[ R_{\text{total}} = r + r + 4 = 2r + 4 \]
Using Ohm's law, the current \( I \) through the resistor can be expressed as:
\[ I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{2E}{2r + 4} \]

### Step 3: Parallel Connection
In a parallel connection, the voltage across the resistor remains the same as the voltage of one cell:
\[ V_{\text{parallel}} = E \]
The effective internal resistance for two identical cells in parallel is given by:
\[ R_{\text{internal}} = \frac{r}{2} \]
Thus, the total resistance in the circuit is:
\[ R_{\text{total}} = \frac{r}{2} + 4 \]
The current \( I \) through the resistor in this case is:
\[ I = \frac{E}{\frac{r}{2} + 4} \]

### Step 4: Equate the Currents
Since the problem states that the current through the resistor is the same in both configurations, we can set the two expressions for current equal to each other:
\[ \frac{2E}{2r + 4} = \frac{E}{\frac{r}{2} + 4} \]

### Step 5: Cross Multiply and Simplify
Cross multiplying gives us:
\[ 2E \left( \frac{r}{2} + 4 \right) = E(2r + 4) \]
Dividing both sides by \( E \) (assuming \( E \neq 0 \)):
\[ 2 \left( \frac{r}{2} + 4 \right) = 2r + 4 \]
Expanding and simplifying:
\[ r + 8 = 2r + 4 \]
Rearranging gives:
\[ 8 - 4 = 2r - r \]
\[ 4 = r \]

### Conclusion
The internal resistance \( r \) of the cell is \( 4 \, \Omega \).