NCERT Question 11 Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
This can be done in 4 ways
All 3 in series
All 3 in parallel
2 in parallel, 1 in series
2 in series, 1 in parallel
In series combination,
equivalent resistance is given by
Rs = R1 + R2 + R3
In parallel combination,
equivalent resistance is given by
1/𝑅_𝑃 = 1/𝑅_1 + 1/𝑅_2 + 1/𝑅_3
Case I All are connected in series
R = 6 + 6 + 6
= 18 Ω
Case II All are connected in parallel
1/𝑅 = 1/6 + 1/6 + 1/6
1/𝑅 = 3/6
1/𝑅 = 1/2
R = 2 Ω
Case III 2 resistors connected in series and 1 parallel to them
First we’ll find equivalent of resistors in series
RS = 6 + 6
= 12 Ω
Now the circuit becomes
Thus, equivalent resistance becomes
1/𝑅 = 1/12 + 1/6
1/𝑅 = (1 + 2)/12
1/𝑅 = 3/12.
1/𝑅 = 1/12
R = 4 Ω
Case IV 2 resistors connected in parallel and 1 connected in series with them
First we’ll find equivalent of resistors in parallel
1/𝑅_𝑝 = 1/6 + 1/6
1/𝑅_𝑝 = 2/6
1/𝑅_𝑝 = 1/3
Rp = 3 Ω
Now the circuit becomes
Thus, equivalent resistance becomes
R = 3 + 6
= 9 Ω
Therefore,
To give total resistance of 9 Ω
Case IV Connect 2 resistors in parallel and 1 resistors in series with them
(b) To give total resistance of 4 Ω
Case III Connect 2 resistors in series and 1 resistors parallel to them

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.