In the above question, if the internal resistance of the battery is 1 ohm , then what is the reading of ammeter

Options

A. `5//3A`
B. `40//29A`
C. `10//9 A`
D. `1A`

Answer

To find the reading of the ammeter when the internal resistance of the battery is 1 ohm, we will follow these steps:

### Step 1: Understand the Circuit
We have a circuit with a 10V battery, a 5-ohm resistor, a 4-ohm resistor, and the internal resistance of the battery is 1 ohm. The total current flowing through the circuit is denoted as \( I \), and the current flowing through the 5-ohm resistor is denoted as \( I_1 \).

### Step 2: Apply Kirchhoff's Voltage Law (KVL) in Loop 1
Using KVL in Loop 1, we can write the equation based on the voltage drops across the resistors:

\[ 10V = 5I_1 + 1I \]

This can be rearranged to form our first equation:

\[ 10 = 5I_1 + I \quad \text{(Equation 1)} \]

### Step 3: Apply KVL in Loop 2
Next, we apply KVL in Loop 2, which includes the 5-ohm and 4-ohm resistors:

\[ 5I_1 - 4(I - I_1) = 0 \]

Expanding this gives:

\[ 5I_1 - 4I + 4I_1 = 0 \]

Combining like terms results in:

\[ 9I_1 = 4I \]

This can be rearranged to form our second equation:

\[ I = \frac{9}{4}I_1 \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 2 into Equation 1
Now, we substitute Equation 2 into Equation 1:

\[ 10 = 5I_1 + \frac{9}{4}I_1 \]

To combine the terms, we convert \( 5I_1 \) to a fraction:

\[ 10 = \frac{20}{4}I_1 + \frac{9}{4}I_1 \]

This simplifies to:

\[ 10 = \frac{29}{4}I_1 \]

### Step 5: Solve for \( I_1 \)
Now, we can solve for \( I_1 \):

\[ I_1 = \frac{10 \times 4}{29} = \frac{40}{29} \text{ A} \]

### Step 6: Find the Ammeter Reading
Since the ammeter measures the total current \( I \), we can now substitute \( I_1 \) back into Equation 2 to find \( I \):

\[ I = \frac{9}{4} \times \frac{40}{29} = \frac{360}{116} = \frac{90}{29} \text{ A} \]

Thus, the reading of the ammeter is:

\[ \text{Ammeter Reading} = \frac{90}{29} \text{ A} \]

### Summary of the Solution
The reading of the ammeter is \( \frac{90}{29} \) A.

---