A cell of emf 2 V and internal resistance `0.1Omega ` is connected to `3.9Omega` external resistance. What will be the potential difference across the terminals of the cell?

A cell of emf 2 V and internal resistance 0.1Ohm is connected to a 20 Ohm resistor .What is the current in the circuit ?

In the circuit show in the figure E,F,G and H are cells of e.m.f. 2,1,3 and 1 V and their internal resistances are 2,1,3 and 1Omega respectively. Calculate the potential difference across the terminals of each of the cells G and H.

A cell of e.m.f. 2 V and negligible internal resistance is connected to a potentiometer wire of resistance 10 Omega and length 4 m. The potential difference per unit length (potential gradient ) of the wire is

A cell of e.m.f. E an internal resistance r is connected across an external resistance R. Plot a graph showing the variation of P.D. across R, verses R.

A resistance R is connected across a cell of emf epsilon and internal resistance r. A potentiometer now measures the potential difference between the terminals of the cell as V. Write the expression for 'r' in terms of epsilon , V and R.

A current of 2.0 A passes through a cell of emf 1.5 V having internal resistance of 0.15 Omega . The potential difference (measured in volt) across both the terminals of the cell is

A uniform wire of resistance 12 Omega is cut into three pieces in the ratio 1 : 2 : 3 and the three pieces are connected to form a triangle. A cell of emf 8 V and internal resistance 1 Omega is connected across the highest of the three resistors. Calculate the current through each part of the circuit.

n cells of e.m.f E and internal resistance r are connected in series with an external resistance R . the current will increase n times that of a single cell, if