A battery of emf 5V and internal resistance `20Omega` is connected to a combination of two resistance `50Omega` and `40Omega` in series. If the p.d. across the `50Omega` resistor is measured by a voltmeter of resistance `1000Omega`, what will be the percentage error?

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the total resistance in the circuit
The total resistance (R_total) in the circuit is the sum of the internal resistance of the battery (r_battery) and the resistances of the two resistors in series (R1 and R2).

Given:
- \( r_{battery} = 20 \, \Omega \)
- \( R_1 = 50 \, \Omega \)
- \( R_2 = 40 \, \Omega \)

\[ R_{total} = r_{battery} + R_1 + R_2 = 20 + 50 + 40 = 110 \, \Omega \]

### Step 2: Calculate the current in the circuit
Using Ohm's law, we can calculate the current (I) flowing through the circuit using the formula:

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

Where:
- \( V = 5 \, V \) (emf of the battery)

\[ I = \frac{5}{110} \approx 0.04545 \, A \]

### Step 3: Calculate the potential difference across the 50Ω resistor
The potential difference (V_R1) across the 50Ω resistor can be calculated using:

\[ V_{R1} = I \times R_1 \]

Substituting the values:

\[ V_{R1} = 0.04545 \times 50 \approx 2.2725 \, V \]

### Step 4: Calculate the resistance of the voltmeter in parallel with the 50Ω resistor
The resistance of the voltmeter (R_voltmeter) is given as 1000Ω. When connected in parallel with the 50Ω resistor, the equivalent resistance (R_eq) can be calculated using the formula:

\[ R_{eq} = \frac{R_1 \times R_{voltmeter}}{R_1 + R_{voltmeter}} \]

Substituting the values:

\[ R_{eq} = \frac{50 \times 1000}{50 + 1000} = \frac{50000}{1050} \approx 47.619 \, \Omega \]

### Step 5: Calculate the new current in the circuit with the voltmeter
The new total resistance in the circuit (R_total_new) will now include the equivalent resistance of the voltmeter:

\[ R_{total\_new} = r_{battery} + R_{eq} + R_2 = 20 + 47.619 + 40 = 107.619 \, \Omega \]

Now, calculate the new current (I_new):

\[ I_{new} = \frac{V}{R_{total\_new}} = \frac{5}{107.619} \approx 0.0465 \, A \]

### Step 6: Calculate the potential difference measured by the voltmeter
The potential difference (V_measured) across the equivalent resistance is:

\[ V_{measured} = I_{new} \times R_{eq} \]

Substituting the values:

\[ V_{measured} = 0.0465 \times 47.619 \approx 2.22 \, V \]

### Step 7: Calculate the percentage error
The percentage error can be calculated using the formula:

\[ \text{Percentage Error} = \left( \frac{V_{true} - V_{measured}}{V_{true}} \right) \times 100 \]

Where \( V_{true} \) is the true value (2.2725 V) and \( V_{measured} \) is the measured value (2.22 V):

\[ \text{Percentage Error} = \left( \frac{2.2725 - 2.22}{2.2725} \right) \times 100 \approx 2.30\% \]

### Final Answer
The percentage error in the measurement is approximately **2.30%**.

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