Question

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of 

1. 9 Ω
2. 4 Ω

Answer 1

(i). To get total 9 Ω resistance from three 6 Ω resistors, we should connect two resistors in parallel and the third resistor in series with the resultant. The combination is given as follows:

Total resistance in parallel is given by
\(\frac {1}{R_{12}} =\frac {1}{𝑅_1} +\frac {1}{R_2}\)

⟹\(\frac {1}{R_{12}} =\frac 16+\frac 16\)

⟹\(\frac {1}{R_{12}}\)\(=\frac 26\)

⟹\(\frac {1}{R_{12}}\)  \(=\frac 13\)

⟹\(R_{12}\)\(=3 \ Ω\)
Now \(𝑅_{12}\) and 6 Ω are connected in series, so the net resistance is given by
\(R=R_{12}+6 Ω\)
\(R =3 \ Ω+6 \ Ω\)
\(R =9\  Ω\)

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(ii) To get total 4 Ω resistance from three 6 Ω resistors, we should connect two resistors in series and the third resistor in parallel with the resultant. The combination is given as follows:

Total resistance in series is given by
\(R_{12}= R_1+R_2\)
\(R_{12} = 6 \ Ω + 6 \ Ω\)
\(R_{12} = 12 \ Ω\)
Now \(R_{12}\) and 6 Ω are connected in parallel, so the net resistance is given by
\(\frac 1𝑅=\frac {1}{R_{12}} +\frac 16\)

⟹\(\frac 1𝑅 =\frac {1}{12}+\frac 16\)

⟹\(\frac 1R =\frac {3}{12}\)

⟹\(\frac 1𝑅 =\frac 14\)

⟹\(𝑅=4 \ Ω\)