Question 8 Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 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 = 9 + 9 + 9
= 27 Ω Case II All are connected in parallel
1/𝑅 = 1/9 + 1/9 + 1/9
1/𝑅 = 3/9
1/𝑅 = 1/3
R = 3 Ω
Case III 2 resistors connected in series and 1 parallel to them
First we’ll find equivalent of resistors in series
RS = 9 + 9
= 18 Ω
Now the circuit becomes
Thus, equivalent resistance becomes
1/𝑅 = 1/18 + 1/9
1/𝑅 = (1 + 2)/18
1/𝑅 = 3/18.
1/𝑅 = 1/6
R = 6 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 = 1/9 + 1/9
1/𝑅_𝑝 = 2/9
Rp = 9/2
Rp = 4.5 Ω Now the circuit becomes
Thus, equivalent resistance becomes
R = 4.5 + 9
= 13.5 Ω Therefore,
To give total resistance of 13.5 Ω
Case IV Connect 2 resistors in parallel and 1 resistors in series with them
(b) To give total resistance of 6 Ω
Case III Connect 2 resistors in series and 1 resistors parallel to them
CA Maninder Singh is a Chartered Accountant for the past 13 years and a teacher from the past 17 years. He teaches Science, Economics, Accounting and English at Teachoo
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