To solve the problem, we need to calculate the potential difference across the 400 Ω resistor when a voltmeter is connected across it. We will also find the error in the measurement due to the presence of the voltmeter.
### Step-by-Step Solution:
1. **Identify the Circuit Configuration**:
- We have two resistors: \( R_1 = 400 \, \Omega \) and \( R_2 = 800 \, \Omega \) connected in series with a 6V battery.
- A voltmeter with a resistance of \( R_v = 10,000 \, \Omega \) is connected across the 400 Ω resistor.
2. **Calculate Total Resistance**:
- The total resistance of the series circuit without the voltmeter is:
\[
R_{total} = R_1 + R_2 = 400 \, \Omega + 800 \, \Omega = 1200 \, \Omega
\]
3. **Calculate the Current in the Circuit**:
- Using Ohm's law, the current \( I \) flowing through the circuit is:
\[
I = \frac{V}{R_{total}} = \frac{6 \, V}{1200 \, \Omega} = 0.005 \, A \, (or \, 5 \, mA)
\]
4. **Calculate the Voltage Across the 400 Ω Resistor (Without Voltmeter)**:
- The voltage across the 400 Ω resistor, \( V_1 \), is given by:
\[
V_1 = I \cdot R_1 = 0.005 \, A \cdot 400 \, \Omega = 2 \, V
\]
5. **Calculate the Equivalent Resistance with the Voltmeter**:
- When the voltmeter is connected, the 400 Ω resistor and the voltmeter are in parallel. The equivalent resistance \( R_{AB} \) across the points A and B (where the voltmeter is connected) is:
\[
R_{AB} = \frac{R_1 \cdot R_v}{R_1 + R_v} = \frac{400 \, \Omega \cdot 10,000 \, \Omega}{400 \, \Omega + 10,000 \, \Omega} = \frac{4,000,000}{10,400} \approx 384.62 \, \Omega
\]
6. **Calculate the New Total Resistance**:
- The new total resistance in the circuit is:
\[
R_{new} = R_{AB} + R_2 = 384.62 \, \Omega + 800 \, \Omega \approx 1184.62 \, \Omega
\]
7. **Calculate the New Current in the Circuit**:
- The new current \( I' \) flowing through the circuit is:
\[
I' = \frac{V}{R_{new}} = \frac{6 \, V}{1184.62 \, \Omega} \approx 0.00506 \, A
\]
8. **Calculate the New Voltage Across the 400 Ω Resistor (With Voltmeter)**:
- The new voltage \( V_1' \) across the 400 Ω resistor is:
\[
V_1' = I' \cdot R_1 = 0.00506 \, A \cdot 400 \, \Omega \approx 2.024 \, V
\]
9. **Calculate the Error in Measurement**:
- The error in measurement \( \Delta V \) is:
\[
\Delta V = V_1 - V_1' = 2 \, V - 2.024 \, V \approx -0.024 \, V
\]
### Final Answer:
The error in the measurement of potential difference is approximately \( -0.024 \, V \).
