Emf and cells in combination

A battery of emf 10V and internal resistance 3Ω is connected to a resistor. The current in the circuit is 0.5A. The tereminal voltage of the battery when the circuit is closed is

• 10V

• zero

• 1.5V

• 8.5V

Two similiar accumulators each of emf E and internal resistance r are connected as shown:

Then potential difference between x and y is

• 2E

• E

• zero

• none of these

To supply maximum current, cells should be arranged in

• series

• parallel

• mixed grouping

• depends on internal and external resistance

The maximum power drawn out of the cell from a source is given by

• ε²/ 2r

• ε²/ 4r

• ε²/ r

• ε²/ 3r

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

• 0.2Ω

• 0.5Ω

• 0.8Ω

• 1.2Ω

The internal resistance of a cell of emf 2V is 0.1Ω.It is connected to a resitance of 3.9Ω.The voltage across the cell will be

• 0.5V

• 1.5V

• 1.9V

• 2V

If n cells of emf E and internal resistance r are connected in series, then current drawn through external resistance R is

• I= nE/ R + nr

• I= E/R+r/n

• I= E/ R + nr

• I= nE/ nR + r

If m cells are connected in parallel, then current drawn through external resistance R is

• I= E/ mR +r

• I= mE/ R +r

• I= mE/ R +mr

• I= mE/ mR +r

A lead acid battery of a car has an emf of 12V. If the internal resistance of the battery is 0.5 ohm, the maximum current that can be drawn from the battery will be

• 30A

• 20A

• 6A

• 24A

If the potential difference across the internal resistance r1 is equal to the emf E of the battery, then

• R= r1 +r2

• R= r1/r2

• R= r1- r2

• R= r2/r1