In a metre bridge with $X = 2\ \Omega$ in the left gap and $R = 3\ \Omega$ in the right gap, the wire (length 1 m, diameter 0.12 cm) has resistance 1.49 Ω. When the bridge is balanced and the cell EMF is 2 V, the current drawn from the cell is approximately:
A0.4 A
B2.0 A
C1.0 A
D1.74 A
Answer & Solution
Correct answer: D. 1.74 A
At balance, $X + R = 5\ \Omega$ acts in parallel with the wire (1.49 Ω). Parallel resistance $R_p = (5\times1.49)/(5+1.49) \approx 1.15\ \Omega$. Current $I = \varepsilon/R_p = 2/1.15 \approx 1.74$ A.
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