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 :

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 :

The galvanometer deflection, when key K₁ is closed but K₂ is open, equals $$\theta $$₀ (see figure). On closing K₂ also and adjusting R₂ to 5$$\Omega $$, the deflection in galvanometer becomes $${{{\theta _0}} \over 5}.$$ . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :

Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P₁ and P₂ respectively, then :