Question

Show with the help of diagrams, how you would connect three resistors each of resistance 6 Ω so that the combination has resistance of (i) 9 Ω (ii) 4 Ω

Answer 1

Each resistor has resistance of 6 Ω.

(i) 1/Rp + Rs

Means two connected parallel and then in series with one.

→ [(1/6 + 1/6) + 6] Ω

→ (3 + 6)Ω

→ 9 Ω

(ii) Rs + Rp

Means two connected in series and then in parallel with 3rd one.

→ (6 + 6) + 1/6

→ 1/12 Ω + 1/6 Ω

→ (2 + 3)/12 Ω

→ 4 Ω

Refer to attachment

Answer 2

» CASE 1.

We have to connect the three resistors such that the combination of the resistance is 9 Ω

• $\dfrac{1}{R_{P_{1} }}$ + $\dfrac{1}{R_{P_{2} }}$ + $R_{S}$

Here .. two connected in parallel and one in series

→ R = $\dfrac{1}{6}$ + $\dfrac{1}{6}$ + 6

→ R = $\dfrac{1\:+\:1}{6}$ + 6

→ R = $\dfrac{2}{6}$ + 6

→ R = 3 + 6 = 9 Ω

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» CASE 2.

We have to connect the three resistors such that the combination of the resistance is 4 Ω.

• $\dfrac{1}{ R_{P}} \: + \: R_{ S_{1} } \: + \:R_{ S_{2} }$

Here .. two resistors are connected in series and in last they both are connected in parallel.

→ $R_{S}$ = 6 + 6

→ $R_{S}$ = 12

→ $\dfrac{1}{R_{P}}$ = $\dfrac{1}{6}$ + $\dfrac{1}{12}$

→ $\dfrac{1}{R}$ = $\dfrac{2\:+\:1}{12}$

→$\dfrac{1}{R}$ = $\dfrac{3}{12}$

→$\dfrac{1}{R}$ = $\dfrac{1}{4}$

→ R = 4 Ω

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