Question

In given circuit, reading of voltmeter is 1 V, then resistance of

• A. 100 Ω
• B. 200 Ω
• C. 200√5 Ω
• D. 50 Ω

Answer 1

The Correct Option is A

This problem requires determining the internal resistance of a voltmeter. The scenario involves a series circuit with a 5 V source, a 200 Ω resistor, and a 100 Ω resistor. A voltmeter is connected in parallel with the 100 Ω resistor, and its reading is provided.

Underlying Principles:

The solution is based on fundamental electric circuit principles:

1. Ohm's Law: \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.

2. Series Circuits: In a series circuit, the total voltage is the sum of individual voltage drops, and the current is uniform throughout.

3. Parallel Circuits: For two resistors \( R_a \) and \( R_b \) in parallel, the equivalent resistance \( R_p \) is calculated as:

\[\frac{1}{R_p} = \frac{1}{R_a} + \frac{1}{R_b} \quad \text{or} \quad R_p = \frac{R_a R_b}{R_a + R_b}\]

A voltmeter, characterized by its internal resistance, is always connected in parallel to measure voltage across a component.

Solution Steps:

Step 1: Define Given Circuit Parameters.

Source voltage, \( V_s = 5 \, \text{V} \).

Resistor 1, \( R_1 = 100 \, \Omega \).

Resistor 2, \( R_2 = 200 \, \Omega \).

Voltmeter reading across \( R_1 \), \( V_m = 1 \, \text{V} \).

Let the voltmeter's resistance be \( R_V \).

Step 2: Calculate Voltage Across the 200 Ω Resistor.

The parallel combination of the voltmeter and the 100 Ω resistor is in series with the 200 Ω resistor. The sum of voltages across these series elements equals the source voltage.

\[V_s = V_m + V_{200}\]

Substituting known values:

\[5 \, \text{V} = 1 \, \text{V} + V_{200}\]

Therefore:

\[V_{200} = 4 \, \text{V}\]

Step 3: Determine the Total Circuit Current.

This total current \( I \) flows through the 200 Ω resistor. Applying Ohm's Law:

\[I = \frac{V_{200}}{R_2} = \frac{4 \, \text{V}}{200 \, \Omega} = 0.02 \, \text{A}\]

Step 4: Compute the Equivalent Resistance of the Parallel Combination.

The voltmeter reading \( V_m = 1 \, \text{V} \) is the voltage across the parallel arrangement of \( R_1 \) and \( R_V \). Let this equivalent resistance be \( R_p \). Using Ohm's Law for this parallel section:

\[V_m = I \times R_p\]

Substituting values:

\[1 \, \text{V} = (0.02 \, \text{A}) \times R_p\]

Solving for \( R_p \):

\[R_p = 50 \, \Omega\]

Step 5: Calculate the Voltmeter's Resistance \( R_V \).

The equivalent parallel resistance \( R_p \) is derived from \( R_1 \) and \( R_V \) in parallel:

\[R_p = \frac{R_1 \times R_V}{R_1 + R_V}\]

Substitute the known values:

\[50 = \frac{100 \times R_V}{100 + R_V}\]

Final Calculation and Outcome:

Solving the equation for \( R_V \):

\[50(100 + R_V) = 100 R_V\]\[5000 + 50 R_V = 100 R_V\]\[5000 = 50 R_V\]\[R_V = 100 \, \Omega\]

The resistance of the voltmeter is determined to be 100 Ω.