Two resistors 400 $$\Omega$$ and 800 $$\Omega$$ are connected in series across a 6V battery. The potential difference measured by a voltmeter of 10 k$$\Omega$$ across 400 $$\Omega$$ resistor is close to:

We have two resistors, one of resistance $$R_{1}=400\;\Omega$$ and another of resistance $$R_{2}=800\;\Omega$$, connected in series to a battery that maintains a potential difference of $$V=6\text{ V}$$.

During the measurement a voltmeter of internal resistance $$R_{v}=10\ \text{k}\Omega=10000\;\Omega$$ is connected across the 400 $$\Omega$$ resistor. Because the voltmeter is always connected in parallel with the element across which we want to know the potential difference, the 400 $$\Omega$$ resistor and the voltmeter together form a parallel combination.

First we calculate the effective (equivalent) resistance $$R_{p}$$ of the parallel combination of $$R_{1}$$ and $$R_{v}$$. For two resistances in parallel the formula is

$$\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{v}}.$$

Substituting the given values,

$$\frac{1}{R_{p}}=\frac{1}{400}+\frac{1}{10000}.$$

We rewrite each term with the same units (ohms) and keep all steps explicit:

$$\frac{1}{R_{p}}=\frac{1}{400}+\frac{1}{10000} =\frac{25}{10000}+\frac{1}{10000} =\frac{26}{10000}.$$

So

$$R_{p}=\frac{10000}{26}\;\Omega.$$

Performing the division,

$$R_{p}=384.615\;\Omega$$ (to three decimal places).

Now this equivalent resistance $$R_{p}$$ is in series with the untouched 800 $$\Omega$$ resistor. Therefore the total resistance $$R_{T}$$ of the circuit seen by the battery becomes

$$R_{T}=R_{p}+R_{2}=384.615\;\Omega+800\;\Omega=1184.615\;\Omega.$$

With the total resistance known, we make use of Ohm’s law, which states $$I=\dfrac{V}{R}$$, to find the current $$I$$ flowing through the series circuit:

$$I=\frac{6\text{ V}}{1184.615\;\Omega}=0.005067\ \text{A}.$$

Converting amperes to milliamperes for clarity,

$$I=0.005067\ \text{A}=5.067\ \text{mA}.$$

The potential difference across the parallel branch (and hence across both the 400 $$\Omega$$ resistor and the voltmeter, since elements in parallel share the same voltage) is obtained again via Ohm’s law:

$$V_{p}=I\,R_{p}=0.005067\ \text{A}\times384.615\;\Omega.$$

Carrying out the multiplication step by step,

$$V_{p}=0.005067\times384.615 =1.9487\ \text{V}.$$

Rounding to two significant figures that match the options provided,

$$V_{p}\approx1.95\ \text{V}.$$

This is the reading that the voltmeter will register.

Hence, the correct answer is Option D.