Question

A galvanometer has a resistance of 50 Ω and it allows maximum current of 5 mA. It can be converted into voltmeter to measure upto 100 V by connecting in series a resistor of resistance

• A. 5975Ω
• B. 20050Ω
• C. 19950Ω
• D. 19500Ω

Answer 1

The Correct Option is C

We use the formula for the total resistance \(R\) of the voltmeter:

\[ R = \frac{V}{I} - R_G \]

Where:

• \(V = 100 \, \text{V}\) is the maximum voltage.
• \(I = 5 \times 10^{-3} \, \text{A}\) is the maximum current.
• \(R_G = 50 \, \Omega\) is the resistance of the galvanometer.

Substituting the values:

\[ R = \frac{100}{5 \times 10^{-3}} - 50 = 20000 - 50 = 19950 \, \Omega \]

Thus, the resistance required is \(19950 \, \Omega\), which corresponds to Option (3).

Answer 2

The problem requires us to find the value of a series resistance needed to convert a given galvanometer into a voltmeter that can measure up to 100 V.

Concept Used:

To convert a galvanometer into a voltmeter, a high resistance, often called a multiplier resistor (\(R_s\)), is connected in series with the galvanometer. The purpose of this series resistor is to limit the current flowing through the galvanometer to its full-scale deflection current (\(I_g\)) when the maximum desired voltage (\(V\)) is applied across the combination.
According to Ohm's law, the total voltage \(V\) across the voltmeter (the series combination of the galvanometer and the resistor) is given by:

\[ V = I_g (R_g + R_s) \]

where:
\(V\) is the maximum voltage the voltmeter can measure.
\(I_g\) is the current for full-scale deflection of the galvanometer.
\(R_g\) is the internal resistance of the galvanometer.
\(R_s\) is the resistance of the series resistor.

Step-by-Step Solution:

Step 1: Identify and list the given parameters from the problem statement.

Resistance of the galvanometer, \(R_g = 50 \, \Omega\).

Maximum current (full-scale deflection current), \(I_g = 5 \, \text{mA} = 5 \times 10^{-3} \, \text{A}\).

The desired maximum voltage measurement range, \(V = 100 \, \text{V}\).

Step 2: State the formula for the series resistance \(R_s\). We rearrange the main formula \(V = I_g (R_g + R_s)\) to solve for \(R_s\).

\[ \frac{V}{I_g} = R_g + R_s \]

Therefore, the required series resistance is:

\[ R_s = \frac{V}{I_g} - R_g \]

Step 3: Substitute the given numerical values into the formula to calculate \(R_s\).

\[ R_s = \frac{100 \, \text{V}}{5 \times 10^{-3} \, \text{A}} - 50 \, \Omega \]

Step 4: Perform the calculation to find the value of \(R_s\).

First, calculate the ratio \(\frac{V}{I_g}\):

\[ \frac{100}{5 \times 10^{-3}} = \frac{100 \times 1000}{5} = 20 \times 1000 = 20000 \, \Omega \]

Now, subtract the galvanometer resistance from this value:

\[ R_s = 20000 \, \Omega - 50 \, \Omega \] \[ R_s = 19950 \, \Omega \]

Hence, the resistance that must be connected in series to convert the galvanometer into a voltmeter that can measure up to 100 V is 19950 Ω.