Question:
If only \(5%\) of the total current is to be passed through galvanometer of resistance \(\text{G}\) , then the resistance of the shunt will be
If only \(5%\) of the total current is to be passed through galvanometer of resistance \(\text{G}\) , then the resistance of the shunt will be
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Smaller current through galvanometer â very small shunt resistance.
Updated On: Apr 26, 2026
- \(\frac{\text{G}}{15}\)
- \(\frac{\text{G}}{17}\)
- \(\frac{\text{G}}{19}\)
- \(\frac{\text{G}}{21}\)
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The Correct Option is C
Solution and Explanation
Concept:
Shunt resistance: \[ S = \frac{G \cdot I_g}{I - I_g} \] Step 1: Given condition. \[ I_g = 5% = 0.05I \] \[ I - I_g = 0.95I \]
Step 2: Substitute. \[ S = \frac{G \cdot 0.05I}{0.95I} \] \[ S = \frac{G \cdot 0.05}{0.95} = \frac{G}{19} \]
Step 3: Conclusion. \[ S = \frac{G}{19} \]
Shunt resistance: \[ S = \frac{G \cdot I_g}{I - I_g} \] Step 1: Given condition. \[ I_g = 5% = 0.05I \] \[ I - I_g = 0.95I \]
Step 2: Substitute. \[ S = \frac{G \cdot 0.05I}{0.95I} \] \[ S = \frac{G \cdot 0.05}{0.95} = \frac{G}{19} \]
Step 3: Conclusion. \[ S = \frac{G}{19} \]
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