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.
Related questions
For a battery of emf ε and internal resistance r driving current I through external resistA galvanometer of resistance G is converted into a voltmeter reading full-scale voltage V A galvanometer of resistance G shows full-scale deflection at current I_g. To convert it iThe relation between current density j, drift velocity v_d, number density n and charge e A wire has a resistance R. It is stretched uniformly so that its length becomes 2L. The neKirchhoff current law at a junction is a statement ofTwo cells of emf ε and internal resistance r each are connected in parallel. The equivalenA potentiometer of length L is used to compare emfs. If the balance lengths for cells of e