the answer is 1.5 Ω
Explanation
the rule to add resistance in parallel is:
[tex]\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+++\frac{1}{R_n}[/tex]
and, the rule to add resistances in serie is
[tex]R_{eq}=R_1+R_2+R_3+...R_n[/tex]
The total resistance of a series circuit is equal to the sum of individual resistances
so
Step 1
add the resistance in parallel :
so.Consider R1 = 3.3 Ω , R2 = 4.5 Ω , R3 = 2.3 Ω , R4 = 3.7 Ω , R5 = 8.7 Ω , and R6 = 7.0 Ω .
hence
left side:
[tex]\begin{gathered} \frac{1}{R_{eq12}}=\frac{1}{R_1}+\frac{1}{R_2} \\ \frac{1}{R_{eq12}}=\frac{1}{3.3}+\frac{1}{4.5} \\ \frac{1}{R_{eq12}}=0.5252 \\ R_{eq12}=\frac{1}{0.5252} \\ R_{eq12}=1.90 \end{gathered}[/tex]
top
[tex]\begin{gathered} \frac{1}{R_{eq345}}=\frac{1}{R_3}+\frac{1}{R_{42}}+\frac{1}{R_5} \\ hence \\ \frac{1}{R_{eq345}}=\frac{1}{2.3}+\frac{1}{3.7}+\frac{1}{8.7} \\ \frac{1}{R_{eq345}}=0.81999 \\ R_{eq345}=\frac{1}{0.8199} \\ R_{eq345}=1.2195 \end{gathered}[/tex]
hence, we have the equivalent circuit
Step 2
b)now, w have resistance (345) and resistance 6 in serie, so we can add using the formula
[tex]\begin{gathered} R_{eq}=R_1+R_2+R_3+...R_n \\ R_{345}a\text{ and R}_5\text{ are in serie , so} \\ R_{3456}=1.2195+7.0 \\ R_{3456}=8.2195 \end{gathered}[/tex]
so, the new circuit would be
Step 3
finally, we have two resistances in parallel, so
.
[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R_{12}}+\frac{1}{R_{3456}} \\ \frac{1}{R_{eq}}=\frac{1}{1.90}+\frac{1}{8.2195} \\ \frac{1}{R_{eq}}=0.6479 \\ isolate\text{ R}_{eq} \\ 1=0.6479\text{ *R}_{eq} \\ divide\text{ both sides by 0.6479} \\ \frac{1}{0.6479}=\frac{0.6479*R_eq}{0.6479} \\ hence \\ R_{eq}=1.543 \\ rounded \\ R_{eq}=1.5 \end{gathered}[/tex]
therefore, the answer is 1.5 Ω
I hope this helps you
