Answer:
(a) 4.0334Ω
(b)parallel
Explanation:
for resistors connected in parallel;
[tex]\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}[/tex]
Req =3.03Ω , R1 =12.18Ω
[tex]\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}[/tex]
[tex]\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}[/tex]
[tex]\frac{1}{R2}=0.2479[/tex]
R2=1/0.2479
R2=4.0334Ω
(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.
Req = R1+R2
