Class 12 Physics
CBSE and UP Board
Current Electricity
Additional Exercise
Chapter-3 Exercise 3.15
NCERT Solutions Class 12 Physics Chapter 3 Question 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. What 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?
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Ans (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 = (nE)/(R+nr)
= (6×2)/(8.5 + 6 x 0.015) = 12 /8.59 =1.39 A
Terminal voltage, V = IR = 1.39 x 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
Ans (b).
After a long use, emf of the secondary cell, E = 1.9 V
Internal resistance of the cell, r = 380 Ω
Hence, maximum current =E/r=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.