Question

In an ammeter, 5% of the main current passes through the galvanometer. If resistance of the galvanometer is G, the resistance of ammeter will be :

• A. \(\frac{G}{200}\)
• B. \(\frac{G}{199}\)
• C. 199 G
• D. 200 G
• E. None of these

Answer 1

Answer: (E)

To solve this problem, we need to determine the resistance of an ammeter given that 5% of the main current passes through the galvanometer whose resistance is \( G \). Let's work through this step-by-step.

1. Understand the ammeter setup:
   • An ammeter is used to measure current flow and generally consists of a galvanometer in parallel with a shunt resistor.
   • The galvanometer is a sensitive instrument and allows only a small fraction of the total current, i.e., 5% in this case, to pass through it.
   • The remaining 95% of the current passes through the shunt resistor, \( R_s \).
2. Formulate the relation between the shunt resistor and the galvanometer:
   • The current division rule for two parallel resistors yields: \[ \text{Current through galvanometer} = \frac{5}{100}I; \; \text{i.e., } I_g = 0.05I \] where \( I \) is the main current.
   • Therefore, 95% of \( I \) passes through \( R_s \): \[ I_s = 0.95I \]
   • Using Ohm's Law in the parallel circuit: \[ I_g = \frac{V}{G} \quad \text{and} \quad I_s = \frac{V}{R_s} \]
   • Setting up the ratio of currents: \[ \frac{I_g}{I_s} = \frac{1}{19} \] \[ \frac{V/G}{V/R_s} = \frac{1}{19} \] which leads to: \[ R_s = \frac{G}{19} \]
3. Calculate the resistance of the ammeter:
   • The total resistance \( R_a \) of the ammeter is then given by the equivalent resistance of the parallel combination of \( G \) and \( R_s \): \[ \frac{1}{R_a} = \frac{1}{G} + \frac{1}{R_s} \] Substitute \( R_s = \frac{G}{19} \): \[ \frac{1}{R_a} = \frac{1}{G} + \frac{19}{G} = \frac{20}{G} \] Therefore: \[ R_a = \frac{G}{20} \] Hence, the ammeter's resistance is \(\frac{G}{20}\), which is not an option in the given choices.

Thus, the correct answer is: None of these.

Answer 2

Given: - 5% of the main current passes through the galvanometer. - The resistance of the galvanometer is \( G \).

Step 1: Calculating the Shunt Resistance \( S \)

The shunt resistance \( S \) is connected in parallel with the galvanometer such that 95% of the main current passes through the shunt. The current division formula for parallel resistances gives:

\[ \frac{I_g}{I} = \frac{S}{S + G} \]

where \( I_g \) is the current through the galvanometer and \( I \) is the total current. Given that:

\[ \frac{I_g}{I} = 0.05 \]

Substituting this value:

\[ 0.05 = \frac{S}{S + G} \]

Rearranging:

\[ 0.05(S + G) = S \] \[ 0.05G = 0.95S \] \[ S = \frac{G}{19} \]

Step 2: Calculating the Resistance of the Ammeter

The resistance of the ammeter \( R_a \) is the equivalent resistance of the galvanometer and the shunt connected in parallel:

\[ \frac{1}{R_a} = \frac{1}{G} + \frac{1}{S} \]

Substituting the value of \( S \):

\[ \frac{1}{R_a} = \frac{1}{G} + \frac{19}{G} = \frac{20}{G} \] \[ R_a = \frac{G}{20} \]

Since the resistance values provided in the options differ from this result, it is possible that additional context or conditions may influence the choice of answer.

Conclusion:

The problem seems to indicate that the correct answer is marked as a bonus question, suggesting that there may be additional considerations or assumptions needed for a precise determination.