Chapter 31 - Fundamentals of Circuits - Exercises and Problems - Page 918: 60

\begin{array}{|c|c|c|} \hline R&I({\;\rm A})& \Delta V ({\;\rm V})\\ \hline 2\;\Omega & 4 & 8 \\ \hline 4\;\Omega & 2 & 8 \\ \hline 12\;\Omega & \frac{2}{3}=0.667& 8 \\ \hline 6\;\Omega & \frac{4}{3}=1.33& 8 \\ \hline 8\;\Omega & 2& 16 \\ \hline \end{array}

First, we need to find the current that comes out from the battery, and to do so we need to find the equivalent resistance of the whole circuit. We can see that the $6$-$\Omega$ and $12$-$\Omega$ are in parallel, and hence their equivalent is $4\;\Omega$. Now this equivalent of $4\;\Omega$ is in series with the $4$-$\Omega$ resistor, and hence their equivalent is $8\;\Omega$. Now we have two $8\;\Omega$s that are in parallel where their equivalent is $4\;\Omega$. Thus, this equivalent of $4\;\Omega$ is in series with the $2$-$\Omega$. Hence, the equivalent resistance of the whole circle is $6\; \Omega$ Applying Ohm's law to find the current of the battery, $$V_B=IR_{eq}$$ $$I=\dfrac{V_B}{R_{eq}}=\dfrac{24}{6}=\bf 4\;\rm A$$ Now from the figures below, the third reduced circuit below, where the current splits to $I_1$ and $I_2$ and both go for $8\;\Omega$ resistors that are in parallel. This means that the current splits two equal parts. Hence, $I_1=I_2=\frac{1}{2}I=\bf 2\;\rm A$ Now it is obvious, from the original circuit below, that $I$ goes through 2-$\Omega$, $\frac{1}{2}I$ goes through 8-$\Omega$ and 4-$\Omega$. Now we need to find the currents that go through the $6\;\Omega$ and the $12\;\Omega$ resistors (where they are in parallel). So, $$6I_3$$ We cans see that $I_1$ splits to $I_3$ and $I_4$ where $$I_1=I_3+I_4\tag 1$$ And the two resistors are in parallel, so $$V_{5\Omega}=V_{12\Omega}$$ $$6I_3=12I_4$$ Hence, $$I_3=2I_4$$ Plug into (1), $$I_1=2I_4+I_4=0.5I$$ Thus, $$I_4=\frac{1}{6}I=\bf \frac{2}{3}\;\rm A$$ and hence, $$I_3=\frac{1}{3}I=\bf \frac{4}{3}\;\rm A$$ Now we can find the potential difference of each resistor, as shown in the table below. \begin{array}{|c|c|c|} \hline R&I({\;\rm A})& \Delta V ({\;\rm V})\\ \hline 2\;\Omega & 4 & 8 \\ \hline 4\;\Omega & 2 & 8 \\ \hline 12\;\Omega & \frac{2}{3}=0.667& 8 \\ \hline 6\;\Omega & \frac{4}{3}=1.33& 8 \\ \hline 8\;\Omega & 2& 16 \\ \hline \end{array}