An ammeter with resistance `R_A` is connected in series with a resistor R, a battery of emf c and internal resistance `r`. The current measured by the ammeter is `I_A`. Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of `I_A, r, R_A` and `R`. Show that more "ideal" the ammeter, the smaller the difference between this current and the current `I_A`. ltbr. (b) If `R = 3.80 Omega, epsilon = 7.50 V` and `r = 0.45 Omega`, find the maximum value of the ammeter resistance `R_A` so that `I_A` is within `99%` of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a maximum value.

Explanation

To solve the problem step by step, we will break it down into parts as per the question.

### Part (a): Finding the current through the circuit when the ammeter is removed

1. **Understanding the Circuit with the Ammeter**:
When the ammeter (with resistance \( R_A \)) is connected in series with the resistor \( R \) and the battery (with EMF \( \epsilon \) and internal resistance \( r \)), the total resistance in the circuit is \( R + R_A + r \).

The current measured by the ammeter, \( I_A \), can be expressed as:
\[ I_A = \frac{\epsilon}{R + R_A + r} \]

2. **Finding the EMF**:
Rearranging the equation for \( \epsilon \):
\[ \epsilon = I_A \times (R + R_A + r) \tag{1} \]

3. **Removing the Ammeter**:
When the ammeter is removed, the circuit now consists only of the battery and the resistor \( R \). The current in this case, denoted as \( I_A' \), can be expressed as:
\[ I_A' = \frac{\epsilon}{R + r} \]

4. **Substituting for EMF**:
Substitute the expression for \( \epsilon \) from equation (1) into the equation for \( I_A' \):
\[ I_A' = \frac{I_A \times (R + R_A + r)}{R + r} \]

5. **Simplifying the Expression**:
This can be simplified as:
\[ I_A' = I_A \times \frac{R + R_A + r}{R + r} \]

6. **Final Expression**:
Thus, the current through the circuit when the ammeter is removed can be expressed as:
\[ I_A' = I_A \times \left(1 + \frac{R_A}{R + r}\right) \]

### Part (b): Finding the maximum value of the ammeter resistance \( R_A \)

1. **Condition for 99% of Current**:
We need to find \( R_A \) such that \( I_A \) is within 99% of \( I_A' \):
\[ I_A = 0.99 I_A' \]

2. **Substituting for \( I_A' \)**:
Substitute the expression for \( I_A' \):
\[ I_A = 0.99 \times I_A \times \left(1 + \frac{R_A}{R + r}\right) \]

3. **Rearranging the Equation**:
Dividing both sides by \( I_A \) (assuming \( I_A \neq 0 \)):
\[ 1 = 0.99 \left(1 + \frac{R_A}{R + r}\right) \]
\[ \frac{1}{0.99} = 1 + \frac{R_A}{R + r} \]
\[ \frac{1}{0.99} - 1 = \frac{R_A}{R + r} \]

4. **Calculating \( R_A \)**:
\[ R_A = (R + r) \left(\frac{1}{0.99} - 1\right) \]

5. **Substituting Values**:
Given \( R = 3.80 \, \Omega \), \( \epsilon = 7.50 \, V \), and \( r = 0.45 \, \Omega \):
\[ R + r = 3.80 + 0.45 = 4.25 \, \Omega \]
\[ R_A = 4.25 \left(\frac{1}{0.99} - 1\right) \approx 4.25 \times (1.0101 - 1) \approx 4.25 \times 0.0101 \approx 0.0429 \, \Omega \]

### Part (c): Explanation of Maximum Value

The value of \( R_A \) calculated represents a maximum value because if \( R_A \) were to increase beyond this point, the current \( I_A \) would drop below 99% of \( I_A' \). This is due to the fact that as \( R_A \) increases, the total resistance in the circuit increases, thereby reducing the current \( I_A \) measured by the ammeter. Hence, the condition \( I_A = 0.99 I_A' \) would no longer hold.