WorksheetsEmf and cells in combination
Total questions: 10
Worksheet time: 6mins
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
