In the above question, the reading of ammeter is `40/x A`. What is the value of `x`?

Explanation

To find the value of \( x \) in the ammeter reading given as \( \frac{40}{x} \) A, we can follow these steps:

### Step 1: Understand the Circuit
The circuit consists of resistances and a battery. We have a 10 V battery, a 10 ohm internal resistance, and two resistors of 4 ohm and 5 ohm. The current \( I \) splits into two branches, one with current \( I_1 \) through the 5 ohm resistor and the other with current \( I - I_1 \) through the 4 ohm resistor.

### Step 2: Apply Kirchhoff's Loop Law
We will apply Kirchhoff's loop law to the two loops in the circuit.

**Loop 1:**
Starting from the 5 ohm resistor:
\[ 5I_1 - 10 + I \cdot 1 = 0 \]
This simplifies to:
\[ 5I_1 + I = 10 \quad \text{(Equation 1)} \]

**Loop 2:**
Starting from the 4 ohm resistor:
\[ 5I_1 - 4(I - I_1) = 0 \]
Expanding this gives:
\[ 5I_1 - 4I + 4I_1 = 0 \]
Combining like terms results in:
\[ 9I_1 - 4I = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations
From Equation 1, we can express \( I \):
\[ I = 10 - 5I_1 \]

Substituting \( I \) into Equation 2:
\[ 9I_1 - 4(10 - 5I_1) = 0 \]
Expanding this gives:
\[ 9I_1 - 40 + 20I_1 = 0 \]
Combining terms results in:
\[ 29I_1 = 40 \]
Thus, we find:
\[ I_1 = \frac{40}{29} \text{ A} \]

### Step 4: Relate to the Ammeter Reading
The ammeter reading is given as \( \frac{40}{x} \) A. Since we have found \( I_1 = \frac{40}{29} \) A, we can set these equal:
\[ \frac{40}{x} = \frac{40}{29} \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:
\[ 40 \cdot 29 = 40 \cdot x \]
Dividing both sides by 40 results in:
\[ x = 29 \]

### Conclusion
The value of \( x \) is \( 29 \).

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