WAEC - Physics (2024 - No. 43)
A galvanometer of full-scale deflection 10 mA has a resistance of 5\(\Omega\). Calculate the shunt needed to convert it to measure currents of 2 A.
0.025\(\Omega\)
0.050\(\Omega\)
0.250\(\Omega\)
0.500\(\Omega\)
Explanation
Voltage across shunt = Voltage across galvanometer
I\(_s\)R\(_s\) = I\(_g\)R\(_g\)
R\(_g\) = 5\(\Omega\), I\(_g\) = 10mA = 0.01A, I\(_s\) = 2 - 0.01 = 1.99A
R\(_s\) = \(\frac{I_g \times R_g}{I_s}\) = \(\frac{0 .01 \times 5}{1.99}\) = 0.025\(\Omega\)
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