Equivalent resistance, source current and a branch current: 22-Elec-A1 Circuits, May 2017, Q1

Question 1: Equivalent resistance, source current and a branch current [10 + 4 + 6]

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. A $100\text{ V}$ source $E_s$ drives the terminal pair a–b. Reading Figure 1: the $5\,\Omega$ is in series from a to the whole top rail (one node, call it $T$). From $T$ three resistors drop to the network: $10\,\Omega$ to node $D$, $12\,\Omega$ to node $E$, and $20\,\Omega$ straight to b. A $6\,\Omega$ bridges $D$–$E$; $11\,\Omega$ joins $D$–b and $15\,\Omega$ joins $E$–b.

Branch	Value	Branch	Value
a–$T$ (series)	$5\,\Omega$	$D$–$E$ (bridge)	$6\,\Omega$
$T$–$D$	$10\,\Omega$	$D$–b	$11\,\Omega$
$T$–$E$	$12\,\Omega$	$E$–b	$15\,\Omega$
$T$–b	$20\,\Omega$	Source $E_s$	$100\text{ V}$

Find. $R_{ab}$, then $I_s=E_s/R_{ab}$, then the current $I_1$ down the $20\,\Omega$ branch.

[Figure not reproduced: Figure 1 — redrawn. The entire top wire right of the $5\,\Omega$ is a single node $T$; the $6\,\Omega$ bridges the two inner columns, so the $T$–$D$–$E$–b block is a bridge that has no pure series/parallel reduction. See the official exam paper.]

Approach. The $D$–$E$–b sub-network is a $\Delta$; convert it to a Y so the remainder collapses by series/parallel, giving $R_{Tb}$ and hence $R_{ab}=5+R_{Tb}$; then $I_s=100/R_{ab}$ and $I_1=V_T/20$ with $V_T=100-5I_s$.

Convert the $\Delta$ ($6,11,15$) on nodes $D,E,$b to a Y. With $\Sigma=6+11+15=32\,\Omega$, the Y arms at $D$, $E$ and b are $$R_D=\frac{6\cdot11}{32}=2.063\,\Omega,\quad R_E=\frac{6\cdot15}{32}=2.813\,\Omega,\quad R_b=\frac{11\cdot15}{32}=5.156\,\Omega.$$
Add the series arms up each column. The Y centre now reaches $T$ through two parallel legs: $10+R_D=12.063\,\Omega$ and $12+R_E=14.813\,\Omega$.
Parallel the two legs, then add $R_b$. $$\frac{12.063\cdot14.813}{12.063+14.813}=6.649\,\Omega,\qquad 6.649+R_b=6.649+5.156=11.806\,\Omega\ (T\text{ to }b\text{ via the Y}).$$
Parallel that with the direct $20\,\Omega$ ($T$–b). $$R_{Tb}=\frac{11.806\cdot20}{11.806+20}=7.423\,\Omega.$$
(a) Add the series $5\,\Omega$. $$\boxed{R_{ab}=5+7.423=12.42\,\Omega.}$$
(b) Source current. $$\boxed{I_s=\frac{E_s}{R_{ab}}=\frac{100}{12.42}=8.05\text{ A}.}$$
(c) Current in the $20\,\Omega$. The top node sits at $V_T=100-5I_s=100-5(8.05)=59.75\text{ V}$ above b, and the $20\,\Omega$ runs straight from $T$ to b, so $$\boxed{I_1=\frac{V_T}{20}=\frac{59.75}{20}=2.99\text{ A}.}$$

Quantity	Value
Equivalent resistance $R_{ab}$	$12.42\,\Omega$
Source current $I_s$	$8.05\text{ A}$
Current in $20\,\Omega$, $I_1$	$2.99\text{ A}$

Topic: Resistive reduction of a bridge network by the $\Delta$–Y transformation.

Key relations: $\Delta\!\to\!$Y: $R_k=\dfrac{(\text{two adjacent }\Delta\text{ arms})}{\sum\Delta}$; series $R=R_1+R_2$; parallel $R=\dfrac{R_1R_2}{R_1+R_2}$; Ohm’s law $V=IR$.

Why this works: A balanced-looking bridge with a resistor across the middle ($6\,\Omega$ here) cannot be peeled apart by series/parallel steps alone, because no two resistors share exactly one node with nothing else attached. Replacing any one $\Delta$ of three resistors by its equivalent Y removes that obstruction — the new star centre introduces the missing "series handle" so the rest folds up. The choice of which $\Delta$ to transform is free; the $D$–$E$–b triangle is convenient because it leaves clean series arms up to $T$.

Common pitfall: treating the top wire as several nodes (it is one node $T$, so $10$, $12$ and $20\,\Omega$ all hang from the same point), or forgetting the $5\,\Omega$ series drop when converting $R_{ab}$ into $I_s$ — $I_1$ uses $V_T$, not the full $100\text{ V}$.

Source: Sadiku, Fundamentals of Electric Circuits, §2.5–2.7 (series/parallel, Wye–Delta).

Question 1 of 6: Equivalent resistance, source current and a branch current

Question 1: Equivalent resistance, source current and a branch current [10 + 4 + 6]