NCERT Solution for Class 10 Science Chapter 12
Electricity
NCERT Books for Session 2022-2023
CBSE Board and UP Board
Intext Questions
Page No-216
Questions No-4
How can three resistors of resistances 2 ohm, 3 ohm, and 6 ohm be connected to give a total resistance of
(a) 4 ohm, (b) 1 ohm?
(a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series
Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R1
Therefore, 1/R1=1/6Ω+1/3Ω=1+2/6Ω=3/6Ω=1/2Ω
Thus R1=2Ω
Now, total effective resistance in the circuit = R1 + 2 Ω = 2 Ω + 2 Ω = 4 Ω
Hence, when resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series, then the total effective resistance in the circuit = 4 Ω
(b) When all the three resistance is connected in parallel then
1/R=1/2Ω+1/3Ω+1/6Ω=3+2+1/6Ω=6/6Ω=1Ω
Thus, R=1Ω
When all the three resistance will be connected in parallel, then the total effective resistance in the circuit = 1 Ω
(i) We can get a total resistance of 4Ω by connecting the 2Ω resistance in series with the parallel combination of 3Ω and 6Ω.
R= R1 + (R2 x R3)/ (R2 + R3)
= 2 + (3 x 6)/ (3+6) = 4 ohm.
(ii) We can obtain a total resistance of 1Ω by connecting resistors of 2 Ω, 3 Ω and 6 Ω in parallel.
1/R = 1/R1 + 1/R2 = 1/2 +1/3+ 1/6 =1 ohm.