NCERT Solution for Class 10 Science Chapter 12
Electricity
NCERT Books for Session 2020-2021
CBSE Board and UP Board
Exercises Questions
Page No-221
Questions No-9
A battery of 9 V is connected in series with resistors of 0.2 ohm, 0.3 ohm, 0.4 ohm , 0.5 ohm and 12 ohm, respectively. How much current would flow through the 12 ohm resistor?
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Total resistance of resistors when connected in series is given by 𝑅=𝑅1+𝑅2+𝑅3+𝑅4+𝑅5 ⟹𝑅=0.2 Ω+0.3 Ω+0.4 Ω+0.5 Ω+12 Ω=13.4 Ω
According to Ohm’s law, V = IR
⟹𝐼=𝑉/𝑅=9/13.4=0.67 𝐴
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Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω
Potential difference, V = 9 V
Current through the series circuit, I = V/R = 12V/13.4Ω = 0.67 A
∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.