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WAEC Physics 2024 Paper

Question : 43

Total: 50

\Omega

Solution:

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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