Question 1 of 6: Equivalent resistance and branch current

Question 1: Equivalent resistance and branch current [10 + 10]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

[Figure not reproduced: Figure‑1 — DC resistive network (redrawn from the printed figure). 30 V source across a–b; I₀ is the current in the right‑hand branch. See the official exam paper.]

Given. A $30\text{ V}$ source drives terminals a–b. Label the interior junctions as redrawn: $C$ (just after the input $10\,\Omega$), $F$ (right of the $25\,\Omega$) and $D$ — the entire bottom rail together with the right-hand return wire, so the top-right node $E$ (top of the $30\,\Omega$) is the same electrical node as $D$. $I_o$ is the current in that right-hand return branch $E\!\to\!D$.

Find. (a) $R_{ab}$; (b) $I_o$.

Approach. The top $40\,\Omega$ lands on node $D$ (through the right-hand wire $E\!\equiv\!D$), so it parallels the $20\,\Omega$; the two $F\!-\!D$ resistors parallel each other; the network then reduces by pure series/parallel and Ohm’s law gives every branch current.

1. Parallel pairs at the ends. The $20\,\Omega$ and the top $40\,\Omega$ both connect $C$ to $D$: $R_{CD}'=20\,\|\,40=\dfrac{20\cdot40}{60}=13.33\,\Omega.$ The $30\,\Omega$ and lower $40\,\Omega$ both connect $F$ to $D$: $R_{FD}=30\,\|\,40=\dfrac{30\cdot40}{70}=17.14\,\Omega.$
2. Series then parallel across $C\!-\!D$. The $25\,\Omega$ in series with $R_{FD}$ forms the second $C\!-\!D$ path, $25+17.14=42.14\,\Omega$, in parallel with $R_{CD}'$: $$R_{CD}=13.33\,\|\,42.14=\frac{13.33\cdot42.14}{55.47}=10.13\,\Omega.$$
3. Equivalent resistance at a–b. Add the input and return resistors in series with $R_{CD}$: $$\boxed{R_{ab}=10+R_{CD}+5=10+10.13+5=25.13\,\Omega.}$$
4. Source and node voltages. The source current is $I_T=\dfrac{30}{R_{ab}}=\dfrac{30}{25.13}=1.194\text{ A}$. With $b=0$, $V_D=I_T(5)=5.97\text{ V}$ and $V_C=30-I_T(10)=18.06\text{ V}$. The current in the $25\,\Omega$ branch is $I_{CF}=\dfrac{V_C-V_D}{42.14}=0.287\text{ A}$, giving $V_F=V_C-25\,I_{CF}=10.89\text{ V}$.
5. Branch current $I_o$. $I_o$ is the current leaving node $E\!\equiv\!D$ up the right wire, which by KCL equals the current arriving there from the top $40\,\Omega$ (from $C$) plus that from the $30\,\Omega$ (from $F$): $$\boxed{I_o=\frac{V_C-V_D}{40}+\frac{V_F-V_D}{30}=\frac{18.06-5.97}{40}+\frac{10.89-5.97}{30}=0.302+0.164=0.466\text{ A}.}$$