Equivalent resistance and branch currents of a bridge network: 22-Elec-A1 Circuits, Undated paper, Q1

Question 1: Equivalent resistance and branch currents of a bridge network [10 + 5 + 5]

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 1 — $110\text{ V}$ source, a $15\,\Omega$ series resistor (current $I_1$), and a Wheatstone bridge: two identical $9\,\Omega$–$5\,\Omega$ legs bridged at their mid-nodes D, E by a $3\,\Omega$ arm (current $I_2$), with a $10\,\Omega$ branch across the same rails.

Given. $V_{dc}=110\text{ V}$ across A(+)–B($-$). A $15\,\Omega$ resistor carries $I_1$ into the top rail (node C). Three branches run C → B: a left leg $9\,\Omega$ then $5\,\Omega$ (mid-node D), an identical middle leg (mid-node E), and a single $10\,\Omega$. A $3\,\Omega$ arm bridges D–E.

Find. (a) $R_{AB}$; (b) $I_1$; (c) $I_2$.

Approach. Test the bridge balance first — if balanced, the $3\,\Omega$ centre arm carries no current and the network collapses to a simple series/parallel ladder.

Check the bridge balance. The two mid-nodes are D and E. The balance condition compares the top-arm/bottom-arm ratios of the two legs: $\dfrac{R_{CD}}{R_{DB}}=\dfrac{9}{5}$ and $\dfrac{R_{CE}}{R_{EB}}=\dfrac{9}{5}$. They are equal, so D and E sit at the same potential and the bridge is balanced. Hence $I_2=0$ — the $3\,\Omega$ arm carries no current and may be removed.
Reduce the three parallel branches. With the bridge open, each outer leg is a plain series pair $9+5=14\,\Omega$, and these parallel the lone $10\,\Omega$:$$R_{CB}=14\,\|\,14\,\|\,10=\left(\tfrac1{14}+\tfrac1{14}+\tfrac1{10}\right)^{-1}=4.118\,\Omega.$$
Add the series input resistor. $R_{AB}=15+R_{CB}=15+4.118=\boxed{19.12\,\Omega}.$
Source / input current. By Ohm’s law at the terminals, $I_1=\dfrac{V_{dc}}{R_{AB}}=\dfrac{110}{19.12}=\boxed{5.75\text{ A}}.$ The voltage across the parallel block is then $V_{CB}=I_1R_{CB}=5.75\times4.118=23.69\text{ V}$.
Bridge current. Because the network is balanced, $\boxed{I_2=0\text{ A}}$ — independent of the $3\,\Omega$ value.

Quantity	Result
Equivalent resistance $R_{AB}$	$19.12\,\Omega$
Series / input current $I_1$	$5.75\text{ A}$
Bridge current $I_2$ (balanced)	$0\text{ A}$

Concept & Theory

Topic: the balanced Wheatstone bridge and series/parallel reduction.

Key relations: balance $\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4}$ gives zero bridge-arm current; parallel combination $\left(\sum 1/R_i\right)^{-1}$; Ohm’s law $I=V/R$.

Why this works: when the arm ratios match, the two mid-nodes are equipotential, so no current crosses the bridging arm whatever its resistance; the bridge then behaves as two independent series legs in parallel. Spotting the symmetry converts an apparently irreducible network (no pure series/parallel step exists while the bridge conducts) into a one-line calculation and removes any need for a Δ–Y transform.

Common pitfall: attempting series/parallel merges before testing balance, or assuming the $3\,\Omega$ value must matter. It does not — only the arm ratios decide the bridge current. Source: Sadiku, Fundamentals of Electric Circuits, 7th ed., §2.6–2.7 and the Wheatstone-bridge example in Ch. 2.

Question 1 of 6: Equivalent resistance and branch currents of a bridge network

Question 1: Equivalent resistance and branch currents of a bridge network [10 + 5 + 5]