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Step-by-step Solution
Step 1: Understand the Circuit
Two resistors, one of $400\,\Omega$ (call it $R_1$) and the other of $800\,\Omega$ (call it $R_2$), are connected in series to a 6 V battery. A voltmeter, which itself has an internal resistance of $10\,\text{k}\Omega$ (call it $R_v$), is placed across the $400\,\Omega$ resistor. This voltmeter is not ideal and therefore must be considered as a resistor in parallel with $R_1$.
Step 2: Combine the Resistor and Voltmeter in Parallel
Since $R_1$ (400 Ω) and $R_v$ ($10\,000\,\Omega$) are in parallel, their effective resistance $R_{\text{parallel}}$ is given by the parallel combination formula:
$$
R_{\text{parallel}} \;=\; \frac{R_1 \times R_v}{R_1 + R_v}
\;=\;
\frac{400 \times 10000}{400 + 10000} \;=\; \frac{4\,000\,000}{10\,400}
\approx 384.615\,\Omega.
$$
Step 3: Calculate the Total Circuit Resistance
The parallel combination $R_{\text{parallel}}$ is now in series with $R_2$ (800 Ω). Hence, the total resistance $R_{\text{total}}$ of the circuit is:
$$
R_{\text{total}} \;=\; R_{\text{parallel}} + R_2
\;=\; 384.615\,\Omega + 800\,\Omega
\;=\; 1184.615\,\Omega \approx 1184.62\,\Omega.
$$
Step 4: Find the Current Through the Circuit
The total current $I$ supplied by the 6 V battery is:
$$
I \;=\; \frac{V_{\text{battery}}}{R_{\text{total}}}
\;=\; \frac{6}{1184.615}
\;\approx\; 0.00507\,\text{A} \;=\; 5.07\,\text{mA}.
$$
Step 5: Determine the Voltmeter Reading
The voltage across the parallel combination (which is the same as the voltmeter reading) is:
$$
V_{\text{across parallel}} \;=\; I \times R_{\text{parallel}}
\;=\; 0.00507 \,\text{A} \times 384.615\,\Omega
\;\approx\; 1.95 \,\text{V}.
$$
Thus, the voltmeter will read approximately 1.95 V.
