1.

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Answer»

ANSWER:

a). Connecting them all in series would give US the highest resistance.

{R}_{eff}\:=\:{R}_{1}\:+\:{R}_{2}\:+\:{R}_{3}R

eff

=R

1

+R

2

+R

3

{R}_{eff}R

eff

= 22Ω

b). Connecting these two resistances in parallel would give us lowest resistance.

1/R = 1/6 + 1/12

1/R = 2+1/12

1/R = 3/12

R = 4Ω



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