Question

In a circuit for finding the resistance of a galvanometer by half deflection method, a $6\, V$ battery and a high resistance of $11 \, k\, \Omega$ are used. The figure of merit of the galvanometer is $60 \, mu A$ division. In the absence of shunt resistance, the galvanometer produces a deflection of $\theta = 9$ divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of $\theta / 2$, is closest to :

• A. $550 \, \Omega $
• B. $220 \, \Omega $
• C. $55 \, \Omega $
• D. $110 \, \Omega $

Answer 1

The Correct Option is D

 To find the shunt resistance that can cause the deflection of the galvanometer to be half, we will use the following approach:

1. Understanding Variables:
   • Given:
     • Voltage of the battery, \(V = 6 \, \text{V}\)
     • High resistance, \(R = 11 \, \text{k}\Omega = 11000 \,\Omega\)
     • Figure of merit (current required for one division deflection), \theta = 9 \, \text{divisions}
   • Unknowns:
     • Shunt resistance, \(S\)
     • Resistance of the galvanometer, \(G\)
2. Equations Involved:
   • Current through the galvanometer without shunt, \(I = K \times \theta = 60 \times 10^{-6} \times 9 \, \text{A}\).
   • Total external resistance in series with the battery and galvanometer, \(R + G\).
   • Voltage through the circuit: \(V = I \times (R + G)\).
   • Applying shunt, the new deflection is half, \(\theta/2 = 4.5 \, \text{divisions}\).
   • With shunt, galvanometer current: \(I' = K \times \theta/2\).
   • Shunt current: I' \times G = I_s \times S.
3. Derive Shunt Resistance:
   • Substituting Equation: \(\frac{I' \times G}{I_s \times S} = 1\).
   • Known:\(I' = \frac{60 \times 10^{-6} \times 9}{2}\), \(I_s = 60 \times 10^{-6} \times 9 - \frac{60 \times 10^{-6} \times 9}{2}\).
   • \(G = \frac{V}{I} - R\ = \frac{6}{60 \times 10^{-6} \times 9} - 11000\).
   • Solve for \(S\)using above values with simplification steps.
4. Solution:
   • After calculations:
     Initial circuit current, \(I = 540 \times 10^{-6} \text{A}\), 
     Current for half-deflection, \(I'/2 = 270 \times 10^{-6} \text{A}\), Total galvanometer resistance without shunt:
     Simplifying,
     Equivalent Shunt Resistance: \(S \approx 110 \, \Omega\).
5. Conclusion:
   • The value of the shunt resistance that can cause the deflection of \(\theta / 2\) is approximately \(110 \, \Omega\), which is the correct answer.

Thus, the answer is \(110 \, \Omega\), which matches the given correct answer.