A voltmeter having a resistance of `1800 Omega` employed to measure the potential difference across a `200 Omega` resistor which is connected to the terminals of a dc power supply having an emf of 50 V and an internal resistance of `20 Omega`. What is the percentage decrease in the potential difference across the `200 Omega` resistor as a result of connecting the voltmeter across it?

Options

A. `1%`
B. `5%`
C. `10%`
D. `25%`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the current flowing through the circuit without the voltmeter.
The total resistance in the circuit is the sum of the internal resistance of the power supply and the resistance of the resistor.

Given:
- EMF (E) = 50 V
- Internal resistance (r) = 20 Ω
- Resistor (R) = 200 Ω

Total resistance (R_total) = R + r = 200 Ω + 20 Ω = 220 Ω

Now, we can calculate the current (I) using Ohm's law:
\[ I = \frac{E}{R_{total}} = \frac{50 \, V}{220 \, \Omega} = \frac{5}{22} \, A \]

### Step 2: Calculate the potential difference across the 200 Ω resistor without the voltmeter.
The voltage across the 200 Ω resistor (V1) can be calculated using:
\[ V1 = E - I \cdot r \]

Substituting the values:
\[ V1 = 50 \, V - \left(\frac{5}{22} \, A \cdot 20 \, \Omega\right) \]

Calculating the second term:
\[ I \cdot r = \frac{5}{22} \cdot 20 = \frac{100}{22} = \frac{50}{11} \, V \]

Now substituting this back into the equation for V1:
\[ V1 = 50 - \frac{50}{11} = \frac{550}{11} - \frac{50}{11} = \frac{500}{11} \, V \approx 45.45 \, V \]

### Step 3: Calculate the equivalent resistance when the voltmeter is connected.
The voltmeter has a resistance of 1800 Ω, and it is connected in parallel with the 200 Ω resistor.

The equivalent resistance (R_eq) of two resistors in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R_v} \]
Where:
- \( R = 200 \, \Omega \)
- \( R_v = 1800 \, \Omega \)

Calculating R_eq:
\[ \frac{1}{R_{eq}} = \frac{1}{200} + \frac{1}{1800} \]
Finding a common denominator:
\[ \frac{1}{R_{eq}} = \frac{9}{1800} + \frac{1}{1800} = \frac{10}{1800} = \frac{1}{180} \]

Thus,
\[ R_{eq} = 180 \, \Omega \]

### Step 4: Calculate the new total resistance with the voltmeter connected.
The new total resistance (R_total_new) is:
\[ R_{total\_new} = R_{eq} + r = 180 \, \Omega + 20 \, \Omega = 200 \, \Omega \]

### Step 5: Calculate the new current flowing through the circuit with the voltmeter connected.
Using Ohm's law again:
\[ I_{new} = \frac{E}{R_{total\_new}} = \frac{50 \, V}{200 \, \Omega} = \frac{1}{4} \, A \]

### Step 6: Calculate the new potential difference across the 200 Ω resistor with the voltmeter connected.
The voltage across the 200 Ω resistor (V2) is given by:
\[ V2 = E - I_{new} \cdot r \]

Calculating:
\[ V2 = 50 \, V - \left(\frac{1}{4} \, A \cdot 20 \, \Omega\right) \]
\[ V2 = 50 - 5 = 45 \, V \]

### Step 7: Calculate the change in potential difference.
The change in voltage (ΔV) is:
\[ \Delta V = V1 - V2 = 45.45 \, V - 45 \, V = 0.45 \, V \]

### Step 8: Calculate the percentage decrease in potential difference.
The percentage decrease is given by:
\[ \text{Percentage decrease} = \left(\frac{\Delta V}{V1}\right) \times 100 \]

Substituting the values:
\[ \text{Percentage decrease} = \left(\frac{0.45}{45.45}\right) \times 100 \approx 0.99\% \]

### Final Answer:
The percentage decrease in the potential difference across the 200 Ω resistor as a result of connecting the voltmeter is approximately **1%**.

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