The emf of a cell is 6 V and internal resistance is `0.5 kOmega`. What will be the reading (in volt) of a voltmeter having an internal resistance of `2.5kOmega`?

Answer

To solve the problem, we need to find the reading of the voltmeter when it is connected to a cell with a given EMF and internal resistance. Here’s a step-by-step solution:

### Step 1: Identify the given values
- EMF of the cell (E) = 6 V
- Internal resistance of the cell (r) = 0.5 kΩ = 500 Ω
- Internal resistance of the voltmeter (R_v) = 2.5 kΩ = 2500 Ω

### Step 2: Calculate the total resistance in the circuit
The total resistance (R_total) in the circuit when the voltmeter is connected is the sum of the internal resistance of the cell and the internal resistance of the voltmeter:
\[ R_{total} = r + R_v = 500 \, \Omega + 2500 \, \Omega = 3000 \, \Omega \]

### Step 3: Calculate the current flowing through the circuit
Using Ohm's law, the current (I) flowing through the circuit can be calculated using the formula:
\[ I = \frac{E}{R_{total}} \]
Substituting the values:
\[ I = \frac{6 \, V}{3000 \, \Omega} = 0.002 \, A = 2 \, mA \]

### Step 4: Calculate the voltage across the voltmeter
The voltage reading on the voltmeter (V_v) can be calculated using Ohm's law again, where the voltage across the voltmeter is given by:
\[ V_v = I \times R_v \]
Substituting the values:
\[ V_v = 0.002 \, A \times 2500 \, \Omega = 5 \, V \]

### Step 5: Conclusion
The reading of the voltmeter when connected to the cell is:
\[ \text{Reading of the voltmeter} = 5 \, V \]

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