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Let's break down the problem step-by-step.

### 1. Expected reading of the meter (Ideal case):
The circuit consists of a resistor \( R = 1.5 \text{ k}\Omega \) (with a 5% uncertainty) and an ideal ammeter connected in series with a DC voltage source \( V_1 = 1.3 \text{ VDC} \).

Since the ammeter is ideal, it has no internal resistance. We can use **Ohm's Law** to find the current:

\[
I = \frac{V}{R}
\]

- \( V_1 = 1.3 \text{ V} \)
- \( R = 1.5 \text{ k}\Omega = 1500 \text{ }\Omega \)

\[
I = \frac{1.3}{1500} = 8.67 \times 10^{-4} \text{ A} = 0.867 \text{ mA}
\]

So, the expected reading of the ammeter is **0.867 mA**.

### 2. Uncertainty in the expected reading:
The resistor has a tolerance of 5%, which means the resistance can vary by 5% of its nominal value. To find the uncertainty in the current, we need to calculate the current for the upper and lower limits of the resistance.

- **Upper limit of resistance:**
  
\[
R_{\text{max}} = 1500 + 0.05 \times 1500 = 1500 + 75 = 1575 \text{ }\Omega
\]

\[
I_{\text{min}} = \frac{1.3}{1575} = 8.25 \times 10^{-4} \text{ A} = 0.825 \text{ mA}
\]

- **Lower limit of resistance:**

\[
R_{\text{min}} = 1500 - 0.05 \times 1500 = 1500 - 75 = 1425 \text{ }\Omega
\]

\[
I_{\text{max}} = \frac{1.3}{1425} = 9.12 \times 10^{-4} \text{ A} = 0.912 \text{ mA}
\]

### 3. Uncertainty calculation:
The uncertainty in the current is the difference between the maximum and minimum values divided by 2:

\[
\text{Uncertainty} = \frac{I_{\text{max}} - I_{\text{min}}}{2} = \frac{0.912 - 0.825}{2} = 0.0435 \text{ mA}
\]

### Final result:
- Expected current: \( 0.867 \text{ mA} \)
- Uncertainty: \( \pm 0.0435 \text{ mA} \)

Would you like further details, or do you have any other questions?

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Here are 5 related questions to expand your understanding:
1. How would the calculation change if the ammeter had some internal resistance?
2. What would happen to the current if the voltage source was increased to 2.0 V?
3. How does the uncertainty in resistance affect the power dissipated in the resistor?
4. Can you explain why an ideal ammeter should have zero resistance?
5. How would you calculate the power dissipated in the circuit?

**Tip:** In any circuit, always start with Ohm's Law (\( V = IR \)) for basic calculations, then add complications like uncertainties or non-ideal components step by step.

Figure 3.5 shows an electrical circuit consisting of a DC voltage source V1, a resistor R, and an ammeter AM connected in series. If R = 1.5 kΩ ± 5% and ammeter AM is an ideal ammeter, calculate the expected reading of the meter for a DC source voltage V1 = 1.3 VDC. Also, calculate the uncertainty in the expected reading due to the uncertainty of the given resistor.

Learn how to calculate the current and its uncertainty in a simple electrical circuit with a 1.5 kΩ resistor and a 1.3 V DC source using Ohm's Law. This guide explains key concepts like uncertainty calculations and is suitable for Grades 10-12.

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