The emf of a cell is 6 V and internal resistance is `0.5 kOmega` The reading of a Voltmeter having an internal resistance of `2.5 kOmega` is

Options

A. 6 V
B. 10 V
C. 5 V
D. 0.5 V

Answer

To find the reading of the voltmeter connected across the terminals of a cell with an EMF of 6 V and an internal resistance of 0.5 kΩ, while the voltmeter has an internal resistance of 2.5 kΩ, we can follow these steps:

### Step 1: Understand the Circuit Configuration
The cell has an EMF (E) of 6 V and an internal resistance (r) of 0.5 kΩ. The voltmeter, which has an internal resistance (R_v) of 2.5 kΩ, is connected in parallel with the internal resistance of the cell.

### Step 2: Calculate the Total Resistance in the Circuit
The total resistance (R_total) seen by the cell can be calculated by adding the internal resistance of the cell and the resistance of the voltmeter in parallel:

\[ \frac{1}{R_{total}} = \frac{1}{r} + \frac{1}{R_v} \]

Substituting the values:

\[ \frac{1}{R_{total}} = \frac{1}{0.5 \, k\Omega} + \frac{1}{2.5 \, k\Omega} \]

Converting kΩ to Ω:

\[ \frac{1}{R_{total}} = \frac{1}{500 \, \Omega} + \frac{1}{2500 \, \Omega} \]

Finding a common denominator (2500):

\[ \frac{1}{R_{total}} = \frac{5}{2500} + \frac{1}{2500} = \frac{6}{2500} \]

Thus,

\[ R_{total} = \frac{2500}{6} \approx 416.67 \, \Omega \]

### Step 3: Calculate the Current in the Circuit
Using Ohm's law, we can find the current (I) flowing through the circuit:

\[ I = \frac{E}{R_{total}} \]

Substituting the values:

\[ I = \frac{6 \, V}{416.67 \, \Omega} \approx 0.0144 \, A = 14.4 \, mA \]

### Step 4: Calculate the Voltage Across the Voltmeter
The voltage across the voltmeter (V_v) can be calculated using Ohm's law:

\[ V_v = I \times R_v \]

Substituting the values:

\[ V_v = 14.4 \, mA \times 2.5 \, k\Omega = 14.4 \times 2.5 \, V = 36 \, V \]

However, this voltage is not correct as we need to consider the voltage drop across the internal resistance of the cell.

### Step 5: Calculate the Voltage Drop Across the Internal Resistance
The voltage drop (V_drop) across the internal resistance of the cell can be calculated as:

\[ V_{drop} = I \times r \]

Substituting the values:

\[ V_{drop} = 14.4 \, mA \times 0.5 \, k\Omega = 14.4 \times 0.5 \, V = 7.2 \, V \]

### Step 6: Calculate the Reading of the Voltmeter
Now, the reading of the voltmeter can be calculated as:

\[ V_{reading} = E - V_{drop} \]

Substituting the values:

\[ V_{reading} = 6 \, V - 7.2 \, V = -1.2 \, V \]

This indicates that the voltmeter reading is not possible under these conditions.

### Final Step: Correct Calculation
Revisiting the calculation, we realize that the voltage across the voltmeter is given by:

\[ V_v = I \times R_v = 14.4 \, mA \times 2.5 \, k\Omega = 36 \, V \]

Thus, the reading of the voltmeter is:

\[ V_v = \frac{6 \, V \times 2.5 \, k\Omega}{(0.5 \, k\Omega + 2.5 \, k\Omega)} = \frac{15 \, V}{3 \, k\Omega} = 5 \, V \]

### Conclusion
The reading of the voltmeter is **5 V**.