The emf of a cell E is 15 V as shown in the figure with an internal resistance of `0.5Omega`. Then the value of the current drawn from the cell 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

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

Two cells E_(1) and E_(2) in the circuit shown in figure, have emfs of 5 V and 9 V and internal resistance of 0.3 Omega and 1.2 Omega respestivley. Calculate the value of current flowing through the resistance of 3 Omega .

(a) Six lead-acid type of secondary cells, each of emf 2.0 V and internal resistance 0.015 Omega are joined in series to provide a supply to a resistance of 8.5 Omega . What are the currents drawn from the supply, and its terminal voltage? (b) A secondary cell after long use has an emf 1.9 V and a large internal resistance of 380 Omega . What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

(a) Six lead-acid type of secondary cells each of emf 2-0 V and internal resistance 0-015Omega are jouned in series to provide a supply to a resistance of 8-5Omega . What are the current drawn from the supply and its terminal voltage ? (b) A secondary cells after long use has an emf of 1-9 V and a large internal resistance of 380 Omega . What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?