Question

show how would you connect 3 resistors of each 6 ohm,s o that rge combination has resistance of 9ohm and 4ohm

Answer 1

Let, the three resisters be $\mathsf{R_1, \ R_2}$ and $\mathsf{R_3}$

$\therefore$

$\mathsf{R_1 \ = \ R_2 \ = \ R_3 \ = \ 6}$ Ω



( i ) To make the total resistance 9 Ω

{ See the attachment 1 }

Two resistors are connected in parallel and one resistor is connected in series.

Now,

Total resistance in parallel combination :

$\mathsf{{\dfrac{1}{R_p}} \ = \ {\dfrac{1}{R_1}} \ + \ {\dfrac{1}{R_2}}}$

$\mathsf{{\dfrac{1}{R_p}} \ = \ {\dfrac{1}{6}} \ + \ {\dfrac{1}{6}}}$

$\mathsf{{\dfrac{1}{R_p}} \ = \ {\dfrac{2}{6}}}$

$\mathsf{{\dfrac{1}{R_p}} \ = \ {\dfrac{1}{3}}}$

$\mathsf{R_p}$ = 3 Ω

Now, one resistor is connected in series.

$\therefore$

$\mathsf{R_T \ = \ R_3 \ + \ R_p}$

$\mathsf{R_T \ = \ 6 + 3}$

$\mathsf{R_T \ = \ 9}$ Ω

Total resistance = 9 Ω




( ii ) To make the total resistance 4 Ω

{ See the attachment 2 }

Two resistors are connected in series.

$\therefore$

$\mathsf{R_s \ = \ R_1 \ + \ R_2}$

$\mathsf{R_s \ = \ 6 + 6}$

$\mathsf{R_s}$ = 12 Ω

Now, $\mathsf{R_s}$ and $\mathsf{R_3}$ are in parallel combination.

$\therefore$

$\mathsf{{\dfrac{1}{R_T}} \ = \ {\dfrac{1}{R_s}} \ + \ {\dfrac{1}{R_3}}}$

$\mathsf{{\dfrac{1}{R_T}} \ = \ {\dfrac{1}{12}} \ + \ {\dfrac{1}{6}}}$

$\mathsf{{\dfrac{1}{R_T}} \ = \ {\dfrac{3}{12}}}$

$\mathsf{{\dfrac{1}{R_T}} \ = \ {\dfrac{1}{4}}}$

$\mathsf{R_T}$ = 4 Ω

Total resistance = 4 Ω

Answer 2

hope it helps you....