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JAMB Physics 2018 Paper

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Question : 83

Total: 87

When a known standard resistor of 2.0 is connected to the 0.0cm end of a metre bridge, the balance point is found to be at 55.0cm

1.10Ω
1.64Ω
2.44Ω
27.50Ω

Solution:

$\frac{R_1}{R_2}$ = $\frac{100 - L_1}{L_1}$

$\frac{R_1}{2}$ = $\frac{45}{55}$

R $_1$ = 2 × $\frac{45}{55}$

= 1.64Ω

this is because the known resistor is connected to 0.00cm end and balances at 55cm

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