A) \[2.5\times {{10}^{-2}}\Omega \]
B) \[1.25\times {{10}^{-3}}\Omega \]
C) \[2.5\times {{10}^{-3}}\Omega \]
D) \[1.25\times {{10}^{-2}}\Omega \]
Correct Answer: D
Solution :
| [d] Formula:\[S={{G}^{*}}I/(I-Ig)\] |
| where S= shunt resistance , (I-Ig)= current through the shunt resistance , I=Current in the circuit. |
| Here, galvanometer resistance, G= 25 ohm ; |
| Ig= 0.001 A , I= 2A |
| \[s=({{25}^{*}}0.0001)/(2-0.001)\] |
| \[=1.25\times {{10}^{-3}}\Omega \] |
| A shunt resistance of\[1.25\times {{10}^{-2}}\Omega \]must be connected to make the galvanometer in the range of 0 to 2A |
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