Ncert Exercises & Solutions

3.15 (a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. Whate are the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

(a) Number of secondary cells, n=6

Emf of each secondary cell, E=2.0 V

Internal resistance of each cell, r=0.015 Ω series resistor is connected to the combination of cells.

Resistance of the resistor, R=8.5 Ω

Current drawn from the supply=I, which is given by the relation,

I = \frac{nE}{R + nr}

= \frac{6 \times 2}{8.5 + 6 \times 0.015}

= \frac{12}{8.59} = 1.39 A

Terminal voltage,

V = IR

= 1.39 \times 8.5 = 11.87 A

Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.

(b) After a long use, emf of the secondary cell, E=1.9 V

Internal resistance of the cell, r=380Ω

Hence, maximum current

= \frac{E}{r} = \frac{1.9}{380} = 0.005 A

Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to start the motor of a car, the cell cannot be used to start a motor.

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