In a mixed grouping of identical cells, five rows are connected in parallel and each row contains 10 cells. This combination sends a current I through an external resistance of `20Omega.` if the emf and internal resistance of each cell is 1.5 V and `1 Omega,` respectively then find the value of I.

Options

A. 5/22 A
B. 15/22 A
C. 3/7 A
D. 5/9 A

Explanation

To solve the problem step by step, we will analyze the configuration of the cells and calculate the current \( I \) flowing through the external resistance.

### Step 1: Understand the Configuration
We have 5 rows of cells connected in parallel, with each row containing 10 cells. Each cell has an EMF (electromotive force) of 1.5 V and an internal resistance of 1 Ω.

### Step 2: Calculate the EMF and Internal Resistance for One Row
For one row of 10 cells connected in series:
- The total EMF \( E_{\text{row}} \) of one row is:
\[ E_{\text{row}} = 10 \times 1.5 \, \text{V} = 15 \, \text{V} \]
- The total internal resistance \( r_{\text{row}} \) of one row is:
\[ r_{\text{row}} = 10 \times 1 \, \Omega = 10 \, \Omega \]

### Step 3: Calculate the Equivalent EMF and Internal Resistance for the Parallel Configuration
Since we have 5 rows connected in parallel:
- The equivalent EMF \( E_{\text{eq}} \) remains the same as that of one row:
\[ E_{\text{eq}} = 15 \, \text{V} \]
- The equivalent internal resistance \( r_{\text{eq}} \) for the 5 rows in parallel is calculated using the formula for resistors in parallel:
\[ \frac{1}{r_{\text{eq}}} = \frac{1}{r_{\text{row}}} + \frac{1}{r_{\text{row}}} + \frac{1}{r_{\text{row}}} + \frac{1}{r_{\text{row}}} + \frac{1}{r_{\text{row}}} = \frac{5}{10} \implies r_{\text{eq}} = \frac{10}{5} = 2 \, \Omega \]

### Step 4: Set Up the Circuit with External Resistance
Now, we have:
- Equivalent EMF \( E_{\text{eq}} = 15 \, \text{V} \)
- Equivalent internal resistance \( r_{\text{eq}} = 2 \, \Omega \)
- External resistance \( R = 20 \, \Omega \)

### Step 5: Apply Ohm's Law to Find the Current \( I \)
Using Ohm's law, the total resistance \( R_{\text{total}} \) in the circuit is:
\[ R_{\text{total}} = r_{\text{eq}} + R = 2 \, \Omega + 20 \, \Omega = 22 \, \Omega \]
Now, we can calculate the current \( I \) using the formula:
\[ I = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{15 \, \text{V}}{22 \, \Omega} \]

### Step 6: Final Calculation
Calculating the current:
\[ I = \frac{15}{22} \, \text{A} \]

Thus, the value of the current \( I \) is:
\[ I = \frac{15}{22} \, \text{A} \]

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