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JAMB Physics 2024 Paper

Question : 71

Total: 125

The value of R in the above circuit to make the galvanometer measure 2A is

Solution:

Given: I $_{g}$ = 50mA = 0.05A, I to be measured = 2A, r = 2 $\Omega$ , I $_{s}$ = I - I $_{g}$ = 2 - 0.05 = 1.95A

Shunt(R) = $\; \frac{I_g}{I_s}$ x r

R = $\; \frac{0.05}{1.95}$ x 10 = 0.2564 $\Omega$

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