Find the minimum number of cell required to produce an electron current of `1.5A` through a resistance of `30 Omega` .Given that the emf internal resistance of each cell are `1.5 V` and `1.0 Omega` respectively

In a mixed grouping of identical cells, five rows are connected in parallel and each row contains 10 cells. This combination sends a current I through an external resistance of 20Omega. if the emf and internal resistance of each cell is 1.5 V and 1 Omega, respectively then find the value of I.

The internal resistance of a 2.1 V cell which gives a current of 0.2A through a resistance of 10 Omega is

For the given circuit, If internal resistance of cell is 1.5 Omega , then

The internal resistance of a 2.1 V cell which gives a current 0.2 A through a resistance of 10 Omega

Current passing through 1 Omega resistance is zero. Then the emf E is

The resistance of the four arms A, B, C and D of a wheatstone bridge as shown in the figure are 10 Omega , 20 Omega , 30 Omega and 60 Omega respectively . The emf and internal resistance of the cell are 15 V and 2.5 Omega respectively . If the galvanometer resistance is 80 Omega then the net current drawn from the cell will be

For the given circuit, if internal resistance of cell is 1.5Omega then V_p-V_q is equal to:

Maximum current produced by a cell of emf 10 V and internal resistance 2 Omega is

A battery of EMF E produces currents 4 A and 3 A when connected to external resistance 1 Omega and 2 Omega respectively. The internal resistance of the battery is

24 cells, each of emf 1.5V and internal resistance is 2Omega connected to 12Omega series external resistance. Then,