A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?



Subscribe to our Youtube Channel - https://you.tube/teachoo
Last updated at Aug. 6, 2019 by Teachoo
Subscribe to our Youtube Channel - https://you.tube/teachoo
Transcript
NCERT Question 9 A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor? Potential difference = V = 9 V The arrangement of resistors is as shown: In series connection, The current through each resistors is same as current through the circuit. To find current through circuit We need to find equivalent resistance Finding equivalent resistance We know that, In series combination, the equivalent resistance is given by R = 𝑅_1 + 𝑅_2 + 𝑅_3 + 𝑅_4 + 𝑅_5 = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω Finding current By Ohm’s Law, V = I R R = 𝑉/𝐼 I = 𝑉/𝑅 I = 9/13.4 I = (9 × 10)/134 I = (9 × 5)/67 I = 45/67 I = 0.67 A Therefore, Current through 12 Ω resistor = Current through the circuit = 0.67 A
NCERT Questions
NCERT Question 2
NCERT Question 3
NCERT Question 4
NCERT Question 5
NCERT Question 6
NCERT Question 7
NCERT Question 8
NCERT Question 9 You are here
NCERT Question 10
NCERT Question 11
NCERT Question 12
NCERT Question 13
NCERT Question 14
NCERT Question 15
NCERT Question 16
NCERT Question 17
NCERT Question 18 (a)
NCERT Question 18 (b)
NCERT Question 18 (c)
NCERT Question 18 (d)
NCERT Question 18 (e)
About the Author