Equivalent resistance and loaded node voltage: 22-Elec-A1 Circuits, December 2016, Q1

Question 1: Equivalent resistance and loaded node voltage [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.

Given. From terminal a a $10\,\Omega$ resistor reaches node C. Three shunt resistors hang from the main row to the bottom rail (terminal b): $20\,\Omega$ at C (this is the resistor whose voltage $V$ is asked), $20\,\Omega$ at D and $10\,\Omega$ at E. Along the row, $10\,\Omega$ joins C–D and $15\,\Omega$ joins D–E, while a $25\,\Omega$ resistor bridges C–E directly along the top. The $25\text{ V}$ source ($+$ at a) drives the terminals.

Element	Value	Between nodes
Series to a	$10\,\Omega$	a–C
Shunt at C ($V$)	$20\,\Omega$	C–b
Row	$10\,\Omega$	C–D
Top bridge	$25\,\Omega$	C–E
Shunt at D	$20\,\Omega$	D–b
Row	$15\,\Omega$	D–E
Shunt at E	$10\,\Omega$	E–b

Find. (a) $R_{ab}$ seen by the source; (b) the voltage $V$ across the $20\,\Omega$ resistor at node C when $25\text{ V}$ is applied.

[Figure not reproduced: Figure 1 — redrawn. The $10\,\Omega$ from a leads to the ladder-plus-bridge network C–D–E; the top $25\,\Omega$ shorts across the two row resistors, so the network is a bridged π and cannot be reduced by simple series/parallel steps alone. See the official exam paper.]

Approach. The C–D–E block is a bridged network (the $25\,\Omega$ bypasses the $10+15\,\Omega$ row), so pure series/parallel collapsing stalls; apply one Δ–Y conversion on the top delta C–D–E, then reduce. For part (b) drive the reduced network with $25\text{ V}$ and use the divider to node C.

Identify the delta. Nodes C, D, E carry a delta of $R_{CD}=10\,\Omega$, $R_{DE}=15\,\Omega$ and $R_{CE}=25\,\Omega$ (the top bridge). Convert to a wye with a new star point N:$$R_{C}=\frac{R_{CD}R_{CE}}{R_{CD}+R_{DE}+R_{CE}},\;R_{D}=\frac{R_{CD}R_{DE}}{\Sigma},\;R_{E}=\frac{R_{DE}R_{CE}}{\Sigma},\quad \Sigma=50\,\Omega.$$
Evaluate the wye arms. $R_C=\dfrac{10\cdot25}{50}=5\,\Omega$, $R_D=\dfrac{10\cdot15}{50}=3\,\Omega$, $R_E=\dfrac{15\cdot25}{50}=7.5\,\Omega$. The star point N now feeds three legs: $R_C$ up to C, $R_D$ down through D’s $20\,\Omega$ to b, and $R_E$ down through E’s $10\,\Omega$ to b.
Collapse the two grounded legs. Leg through D: $R_D+20=23\,\Omega$; leg through E: $R_E+10=17.5\,\Omega$. These two run in parallel from N to b: $$23\,\|\,17.5=\frac{23\cdot17.5}{40.5}=9.938\,\Omega.$$
Series up to node C. Add $R_C$: $5+9.938=14.938\,\Omega$ from C to b. This is in parallel with the $20\,\Omega$ shunt at C: $$20\,\|\,14.938=\frac{20\cdot14.938}{34.938}=8.551\,\Omega.$$
Add the series $10\,\Omega$ from a. $$R_{ab}=10+8.551=\boxed{18.55\,\Omega}.$$
(b) Node C voltage under $25\text{ V}$. The source current is $I=25/R_{ab}=1.348\text{ A}$. The voltage across the C–b combination (which equals $V$ across the $20\,\Omega$) is what remains after the series $10\,\Omega$ drop: $$V=25-I\cdot10=25-13.48=\boxed{11.52\text{ V}}.$$ Equivalently $V=25\cdot\dfrac{8.551}{18.55}=11.52\text{ V}$ ✓.

Quantity	Value
Terminal resistance $R_{ab}$	$18.55\,\Omega$
Source current $I$	$1.348\text{ A}$
Voltage across $20\,\Omega$ at C, $V$	$11.52\text{ V}$

Topic: Bridged-network reduction by Δ–Y transformation.

Key relations: $R_C=\tfrac{R_{CD}R_{CE}}{\Sigma}$ (and cyclic), $\Sigma=R_{CD}+R_{DE}+R_{CE}$; parallel pair $R_1\|R_2=\tfrac{R_1R_2}{R_1+R_2}$; divider $V=E\,R_x/R_{ab}$.

Why this works: The top $25\,\Omega$ ties C to E in parallel with the $10+15\,\Omega$ row, so no resistor is purely in series or parallel with another — the classic bridge situation. Replacing the C–D–E delta by an equivalent wye introduces a neutral node that is reducible: the two grounded legs then simply parallel, and the rest peels off in series/parallel order.

Common pitfall: trying to call the $10\,\Omega$ (C–D) and $15\,\Omega$ (D–E) “in series” and ignoring the $25\,\Omega$ bridge — that mis-reads the topology and gives the wrong $R_{ab}$. The delta must be converted first.

Source: Sadiku, Fundamentals of Electric Circuits, §2.7 (wye–delta transformations) and §2.5–2.6 (series/parallel, dividers).

Question 1 of 6: Equivalent resistance and loaded node voltage

Question 1: Equivalent resistance and loaded node voltage [10 + 10]