Question

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

Answer 1

a)

Given: The emf of the secondary cell is 2.0 V, the numbers of the secondary cells are 6 the internal resistance of each cell is 0.015 Ω and the resistance of the resistor is 8.5 Ω.

The current drawn is given as,

I= nE R+nr

Where, the emf of the secondary cell is E, the numbers of the secondary cells are n the internal resistance of each cell is r and the resistance of the resistor is R.

By substituting the given values in the above formula, we get

I= 6×2 8.5+6×0.015 I= 12 8.59 I=1.39 A

The terminal voltage is given as,

V=IR

Where, the terminal voltage is V.

By substituting the given values in the above formula, we get

V=IR =1.39×8.5 =11.87 V

Thus, terminal voltage is 11.87 Vand the current drawn from the supply is 1.39 A.

b)

Given: The emf of the secondary cell is 1.9 V and the internal resistance of the cell is 380 Ω.

The maximum current is given as,

I m = E r

Where, the emf of the secondary cell is E, the maximum current is I m and the internal resistance of the cell is r.

By substituting the given values in the above formula we get.

I m = 1.9 380 =0.005 A

Thus, the maximum current drawn from the cell is 0.005 A.

As a large current is required to start the motor of a car therefore the cell cannot be used to start a motor.