# 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. What are the current drawn from the supply and its terminal voltage?

Given,

There are 6 secondary cells.

Emf of each cell, E = 2 V (In series)

The internal resistance of each cell, r = 0.015 Ω (In series)

And the resistance of the resistor, R = 8.5 Ω

Let I be the current drawn in the circuit.

$\dpi{100} I = \frac{nE}{R + nr}$

$\dpi{100} \implies I = \frac{6(2)}{8.5 + 6(0.015)} = \frac{12}{8.59}$

$\dpi{100} \implies$ I = 1.4 A

Hence current drawn from supply is 1.4 A

Therefore, terminal voltage, V = IR = 1.4 x 8.5 = 11.9 V

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