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
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Correct Answer: Option B
Explanation:
\(\frac{V_1}{V_2}\) = \(\frac{L_1}{(V_2 − L_1)}\)
 \(\frac{R_1}{R_2}\) = \(\frac{ L_1}{(V_2 − L_1)}\)
 \(\frac{2}{R_2}\) = \(\frac{55}{45}\)
 R\(_2\) = 2 × \(\frac{45}{55}\)
= 1.64Ω
\(\frac{V_1}{V_2}\) = \(\frac{L_1}{(V_2 − L_1)}\)
 \(\frac{R_1}{R_2}\) = \(\frac{ L_1}{(V_2 − L_1)}\)
 \(\frac{2}{R_2}\) = \(\frac{55}{45}\)
 R\(_2\) = 2 × \(\frac{45}{55}\)
= 1.64Ω