The emf of a battery is 4.0 V and its internal resistance is `1.5 Omega`. Its potential difference is measured by a voltmeter of resistance `1000 Omega`. Calculate the percentage error in the reading of emf shown by voltmeter.

Answer

To solve the problem, we need to calculate the percentage error in the reading of the emf shown by the voltmeter. Here are the steps to arrive at the solution:

### Step 1: Identify the given values
- Emf of the battery (E) = 4.0 V
- Internal resistance of the battery (r) = 1.5 Ω
- Resistance of the voltmeter (R) = 1000 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 battery and the resistance of the voltmeter:
\[ R_{\text{total}} = r + R = 1.5 \, \Omega + 1000 \, \Omega = 1001.5 \, \Omega \]

### 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:
\[ I = \frac{E}{R_{\text{total}}} = \frac{4.0 \, V}{1001.5 \, \Omega} \]

Calculating this gives:
\[ I \approx 0.003996 \, A \quad (\text{or } 3.996 \, mA) \]

### Step 4: Calculate the potential difference across the voltmeter
The potential difference (V') across the voltmeter can be calculated using Ohm's law:
\[ V' = I \times R = I \times 1000 \, \Omega \]
Substituting the value of I:
\[ V' = 0.003996 \, A \times 1000 \, \Omega = 3.996 \, V \]

### Step 5: Calculate the percentage error
The percentage error in the reading of the emf shown by the voltmeter can be calculated using the formula:
\[ \text{Percentage Error} = \left( \frac{\text{True Value} - \text{Measured Value}}{\text{True Value}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Error} = \left( \frac{4.0 \, V - 3.996 \, V}{4.0 \, V} \right) \times 100 \]
Calculating this gives:
\[ \text{Percentage Error} = \left( \frac{0.004 \, V}{4.0 \, V} \right) \times 100 \approx 0.1\% \]

### Final Result
The percentage error in the reading of emf shown by the voltmeter is approximately **0.1%**.

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