Test Index

NCERT Class XII Chapter
Current Electricity
Questions With Solutions

© examsnet.com
Question : 15 of 24
Marks: +1, -0
(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance of 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 ?
Solution:  
(a) Six cells are joined in series.
Equivalent emf is 2 × 6 = 12 V
Equivalent internal resistance is 0.015 × 6 = 0.09 Ω
Current drawn from supply
I = 128.5+0.09\frac{12}{8.5+0.09} = 128.59\frac{12}{8.59} = 1.39 A
Terminal voltage of battery
V = E – Ir = 12 – 1.39 × 0.09 or V = 12 – 0.1251 = 11.8 V
(b) Maximum current is drawn from a battery when external resistance is treated to be zero.
ImaxI_{max} = 1.9380\frac{1.9}{380} = 0.005 A
To start a car, a current of the order of 100 A is needed, so the battery mentioned above can not drive the starting motor.
© examsnet.com
Go to Question:
Prev Question
Next Question