Two cells `E_1` and `E_2` have and emf of 5V and 9V and internal resistances of 0.3`Omega` and 1.2`Omega` respectively. Calculate the value of current following through the resistance of 3`Omega`.

A few day cells have an emf of 1.5V and an internal resistance of 0.5 Omega each. How many of these cells connected in series would produce a current of 0.6A through a resistance of 10 Omega in the external circuit.

To the parallel combination of two resistances 3 Omega and 1 Omega , a series combination of resistances 2.15 Omega and 1 Omega and a battery are connected. The interval resistance of the battery is 0.1 Omega and the emf is 2V. Determine the values of current flowing through the resistances. Draw the diagram of the circuit.

In the circuit shown the current through 2 Omega resistance is

Two cells having emf l.5Vand 2V have internal resistance 0.2 Omega and 0.3 Omega respectively. What is the equivalent emf and equivalent resistance if they are connected in paralleled combinations?

The equivalent resistance of two coils connected in series and in parallel are 12Omega and 5/3 Omega respectively. Calculate the value of each resistance.

When a resistance of 12 Omega is conencted with a cell of emf 1.5 V,0.1 A current flows through the resistance internal resistance of the cell is

If two resistance are connected in series and parallel give equivalent resistance 12Omega and 5/2 Omega respectively then find their value.

When a resistance of 12 Omega is connected with a cell of emf 1.5 V, 0.1A current flows through the resistance. Internal resistance of the cell is—

A circuit consists of three batteries of emf E_1 = 1 V, E_2 = 2 V and E_3 = 3 V and internal resistances 1Omega, 2 Omega and 1 Omega respectively which are connected in parallel as shown in the figure. The potential difference between points P and Q is

Two cells such of 0.5Omega and e.m.f. of 1.5V are connected in series to a galvanometer having 8Omega resistance. Calculate the current passing through the circuit.