# Six lead-acid types 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. - Physics

Numerical

Six lead-acid types 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?

#### Solution

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 xx 2)/(8.5 + 6 xx 0.015)

= 12/8.59

= 1.39 A

Terminal voltage, V = IR = 1.39 × 8.5 = 11.87 A

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

Concept: Cells, Emf, Internal Resistance
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 3 Current Electricity
Exercise | Q 3.15 (a) | Page 129
NCERT Class 12 Physics Textbook
Chapter 3 Current Electricity
Exercise | Q 15.1 | Page 129

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