To determine the internal resistance of a cell by using a potentiometer, the null point is at $$1 \mathrm{~m}$$ when shunted by $$3 \Omega$$ resistance and at a length $$1.5 \mathrm{~m}$$, when cell is shunted by $$6 \Omega$$ resistance The internal resistance of the cell is

In the given circuit, the current in 8$$\Omega$$ resistance is 1.5 A. The total current (I) flowing in the circuit is

A balanced bridge is shown in the circuit diagram. The metre bridge wire has resistance $$1 \Omega \mathrm{m}^{-1}$$. The current drawn from the battery is (Internal resistance of battery is negligible)

In potentiometer experiment, cells of e.m.f. '$$E_1$$' and '$$E_2$$' are connected in series $$\left(E_1>E_2\right)$$ the balancing length is $$64 \mathrm{~cm}$$ of the wire. If the polarity of $$E_2$$ is reversed, the balancing length becomes $$32 \mathrm{~cm}$$. The ratio $$\frac{E_1}{E_2}$$ is