R3 and R4 are in series, so:
[tex]R_{eq}1=R3+R4=5+10=15[/tex]
R6 and R7 are also in series, so:
[tex]R_{eq}2=R6+R7=8+7=15[/tex]
R5, Req1 and Req2 are in parallel, so:
[tex]\begin{gathered} \frac{1}{R_{eq}3}=\frac{1}{R_{eq1}}+\frac{1}{R_{eq}2}+\frac{1}{R5} \\ \frac{1}{R_{eq}3}=\frac{1}{15}+\frac{1}{15}+\frac{1}{15} \\ \frac{1}{R_{eq}3}=\frac{3}{15} \\ \frac{1}{R_{eq}3}=\frac{1}{5} \\ R_{eq}3=5 \end{gathered}[/tex]
Now, Req 3, R10 and R11 are in series, so:
[tex]R_{eq}4=R_{eq}3+R10+R11=5+10+15=30[/tex]
Now, we can find the total current in that segment using ohm law:
[tex]\begin{gathered} V=IR \\ I=\frac{V}{R} \\ I1=\frac{V_T}{R_{eq}4}=\frac{240}{30} \\ I1=8A \end{gathered}[/tex]
Now, let's find the voltage and the current for R3,R4, R5, R6, R7, R10 and R11
[tex]\begin{gathered} V1=I1\cdot R_{eq}3=8\cdot5=40V \\ \end{gathered}[/tex]
With this value, let's complete one portion of the table:
---------------
Now, R8 and R9 are in series, so:
[tex]R_{eq}5=R8+R9=20+40=60[/tex]
Req5 and Req4 are in parallel, so:
[tex]\begin{gathered} \frac{1}{R_{eq}6}=\frac{1}{R_{eq}4}+\frac{1}{R_{eq}5} \\ \frac{1}{R_{eq}6}=\frac{1}{30}+\frac{1}{60} \\ \frac{1}{R_{eq}6}=\frac{1}{20} \\ R_{eq}6=\frac{20}{1} \\ R_{eq}6=20 \end{gathered}[/tex]
R1 and R2 are in parallel, R12 and R13 are in parallel too, so:
[tex]\begin{gathered} \frac{1}{R_{eq}7}=\frac{1}{R1}+\frac{1}{R2} \\ \frac{1}{R_{eq}7}=\frac{1}{10}+\frac{1}{10} \\ \frac{1}{R_{eq}7}=\frac{2}{10} \\ R_{eq}7=5 \end{gathered}[/tex][tex]\begin{gathered} \frac{1}{R_{eq}8}=\frac{1}{R12}+\frac{1}{R13} \\ \frac{1}{R_{eq}8}=\frac{1}{12}+\frac{1}{60} \\ \frac{1}{R_{eq}8}=\frac{1}{10} \\ R_{eq}8=10 \end{gathered}[/tex]
Therefore, the total resistance of the circuit since Req6, Req 7, Req8 and R14 are in series, is:
[tex]\begin{gathered} R_T=R_{eq}6+R_{eq}7+R_{eq}8+R14=20+5+10+5 \\ R_T=40 \end{gathered}[/tex]
And the total current is:
[tex]I_T=\frac{V_T}{R_T}=\frac{240}{40}=6_{}[/tex]
Answer:
