For the circuit shown, with R₁ = 1.0W, R₂ = 2.0 W, E₁ = 2 V and E₂ = E₃ = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V):

A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 $$ \times $$ 10^–4 A passes through it, its needle ( pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of :

In the given circuit diagram, the currents, I₁ = – 0.3 A, I₄ = 0.8 A and I₅ = 0.4 A, are flowing as shown. The currents I₂, I₃ and I₆, respectively, are :

In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that, the variation $${{dR} \over {d\ell }}$$ of its resistance R with length $$\ell $$ is $${{dR} \over {d\ell }}$$ $$ \propto $$ $${1 \over {\sqrt \ell }}$$. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP ?