Question

26 Show how would you join three resistors, each of resistance 9 Ω so that equivalent resistance of the combination is (a) 3.5 Ω, (b) 6 Ω?

Answer 1

Explanation:

13.5 comes from 9 ohm resistor in series+ 2 x 9 ohm

Eqjvalent esistance = 9+4.5 = 13.5 ohms

6 ohm

2 x 9 resistors in serie= 18 ohms + in parallel with third 9 ohm resistor…

equivalent Reistance = 1/(1/9+1/18) = 1/(3/18) = 6 ohms

hope it was hellful

Answer 2

CORRECT QUESTION :

26 Show how would you join three resistors, each of resistance 9 Ω so that equivalent resistance of the combination is (a) 13.5 Ω, (b) 6 Ω?

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Three \: resistors = 9\Omega \: \: (each) \\ \\ \red{\underline{\bold{To \:Show:}}} \\ \tt: \implies R_{equivalent} = 3.5\Omega \\ \\ \tt: \implies R_{equivalent} = 6\Omega$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \frac{1}{R_{equivalent}} = \frac{1}{R_{1} } + \frac{1}{ R_{2} } \\ \\ \tt: \implies R_{equivalent} = \frac{ R_{1} R_{2} }{ R_{1} + R_{2} } \\ \\ \tt: \implies R_{equivalent} = \frac{9 \times 9}{9 + 9} \\ \\ \tt: \implies R_{equivalent} = \frac{81}{18} \\ \\ \tt: \implies R_{equivalent} =4.5\Omega \\ \\ \bold{Again : } \\ \tt: \implies R_{AB} = R_{equivalent} + R_{3} \\ \\ \tt: \implies R_{AB} =4.5 + 9\\\\ \green{\tt: \implies R_{AB} =13.5 \Omega}$

$\bold{For\:second\:part : } \\ \tt: \implies R_{equivalent} = R_{1} + R_{2} \\ \\ \tt: \implies R_{equivalent} =9 + 9 \\ \\ \tt: \implies R_{equivalent} =18\Omega \\ \\ \bold{Again : } \\ \tt: \implies \frac{1}{ R_{AB}} = \frac{1}{ R_{equivalent} } + \frac{1}{ R_{3} } \\ \\ \tt: \implies \frac{1}{ R_{AB} } = \frac{1}{18} + \frac{1}{9} \\ \\ \tt: \implies R_{AB} = \frac{18}{3} \\ \\ \green{\tt: \implies R_{AB} = 6\Omega }$