A `100 V` voltmeter of internal resistance `20 k Omega` in series with a high resistance `R` is connected to a `110 V` line. The voltmeter reads `5 V`, the value of `R` is

Options

A. `210kOmega`
B. `315kOmega`
C. `420kOmega`
D. `440kOmega`

Answer

To solve the problem step by step, we will analyze the circuit involving the voltmeter and the external resistance \( R \).

### Step 1: Understand the Circuit
We have a voltmeter with an internal resistance \( r = 20 \, k\Omega \) connected in series with a high resistance \( R \). The entire setup is connected to a \( 110 \, V \) power supply. The voltmeter reads \( 5 \, V \).

### Step 2: Calculate the Current Through the Circuit
The current \( I \) flowing through the circuit can be calculated using Ohm's law. The voltage across the voltmeter is given as \( 5 \, V \). The formula for current is:

\[ I = \frac{V_{AB}}{r} \]

Substituting the known values:

\[ I = \frac{5 \, V}{20 \, k\Omega} = \frac{5}{20 \times 10^3} = 2.5 \times 10^{-4} \, A \]

### Step 3: Apply Kirchhoff's Voltage Law
According to Kirchhoff's Voltage Law, the sum of the potential differences in a closed loop must equal zero. The total voltage supplied by the battery is \( 110 \, V \), which is equal to the sum of the voltage drops across the internal resistance of the voltmeter and the external resistance \( R \).

The equation can be written as:

\[ V_{PQ} = I \cdot (r + R) = 110 \, V \]

### Step 4: Substitute Known Values
Substituting the current \( I \) and the internal resistance \( r \):

\[ 2.5 \times 10^{-4} \cdot (20 \times 10^3 + R) = 110 \]

### Step 5: Simplify the Equation
Now, we can simplify the equation:

\[ 2.5 \times 10^{-4} \cdot (20000 + R) = 110 \]

Dividing both sides by \( 2.5 \times 10^{-4} \):

\[ 20000 + R = \frac{110}{2.5 \times 10^{-4}} = 440000 \]

### Step 6: Solve for \( R \)
Now, isolate \( R \):

\[ R = 440000 - 20000 = 420000 \, \Omega \]

### Step 7: Convert to Kilo Ohms
Finally, convert \( R \) to kilo ohms:

\[ R = 420 \, k\Omega \]

### Conclusion
The value of the external resistance \( R \) is \( 420 \, k\Omega \).