An electrical circuit is shown in fig the values of resistances and the directions of the currents are shown A voltmeter of resistance \(400 \Omega\) is connected across the \(400 \Omega\) resistor. The battery has negligible internal resistance.The residing of the voltmeter is. \((\mathrm{A})(10 / 3) \mathrm{V}\) (B) \(5 \mathrm{~V}\) (C) \((20 / 3) \mathrm{V}\) (D) \(4 \mathrm{~V}\)

In the given circuit, we have resistance values and directions of currents. Let's call the \(400 \Omega\) resistor connected with the voltmeter as \(R_1\), and the other resistor as \(R_2\). The resistance of the voltmeter is also \(400 \Omega\). Let the current passing through \(R_1\) be \(I_1\) and that passing through \(R_2\) be \(I_2\).

In a voltage divider circuit, the voltage across the resistors is proportional to the value of the resistors. In our case, we have a voltmeter of resistance \(400 \Omega\) connected across \(R_1\), which is also \(400 \Omega\). Thus, the voltage divider ratio is equal for both parts of the circuit, as the ratio of \(R_1\) and the voltmeter resistance is 1:1.

Since the voltmeter is connected in parallel with \(R_1\), we need to find the combined resistance of this part of the circuit. To do that, we use the formula for parallel resistances: \[\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_V}\] Here, \(R_P\) is the combined parallel resistance, \(R_1 = 400 \Omega\) and \(R_V = 400 \Omega\). Plugging in the values, we get: \[\frac{1}{R_P} = \frac{1}{400} + \frac{1}{400}\] \[R_P = \frac{400 \times 400}{400 + 400} = 200 \Omega\]

Now, we can calculate the total resistance in the circuit by adding the parallel resistance \(R_P = 200 \Omega\) and \(R_2 = 200 \Omega\), which are connected in series: \[R_T = R_P + R_2 = 200 + 200 = 400\Omega\]

Using Ohm's Law, we can calculate the total current in the circuit. Let the total current be \(I_T\), and assume the total voltage provided by the battery is \(V\): \[I_T = \frac{V}{R_T}\] Let's find the relationship between \(I_1\) and \(I_2\) using the given resistances and currents values: \[I_1 R_1 = I_2 R_2\] \[I_1 = I_2\] Now we can express the total current as the sum of \(I_1\) and \(I_2\): \[I_T = I_1 + I_2 = 2I_1\]

As mentioned in Step 2, since both \(R_1\) and the voltmeter have the same resistance values, the voltage across them must also be equal. So let the voltage across the voltmeter be \(V_M\). Then, using Ohm's Law, we can write: \[V_M = R_1 I_1\] Substituting the relationship between currents, we get: \[V_M = R_1 (\frac{I_T}{2})\] As \(I_T = \frac{V}{R_T}\), we can plug it in to get: \[V_M = R_1 (\frac{V}{2R_T}) = 400 \cdot \frac{V}{2 \cdot 400}\] Finally, we have the voltmeter reading: \[V_M = \frac{V}{2}\] Out of the given options, the only one that fits this relationship is \((\mathrm{B})\) \(5 \mathrm{~V}\). This implies that the total voltage provided by the battery is \(10V\), and the voltmeter reads half of it, which is \(5V\).