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 :

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] :

An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and esistance 5 $$\Omega $$. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is :

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 ?