Bridge network — source current: 22-Elec-A1 Circuits, December 2015, Q1

Question 1: Bridge network — source current [20]

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 $50\text{ V}$ DC source feeds a $3\,\text{k}\Omega$ series resistor (carrying $I$) into a bridge network. The bridge has upper arms $6\,\text{k}\Omega$ (left) and $6\,\text{k}\Omega$ (right), a $6\,\text{k}\Omega$ bridging resistor between the two mid-nodes $a$ and $b$, and lower arms $2\,\text{k}\Omega$ (left) and $2\,\text{k}\Omega$ (right).

Find. The current $I$ delivered by the source through the $3\,\text{k}\Omega$ resistor.

[Figure not reproduced: Figure 1 — redrawn bridge. Top node fed through $3\,\text{k}\Omega$; nodes $a$ and $b$ are the bridge mid-points joined by the $6\,\text{k}\Omega$ bridging resistor. See the official exam paper.]

Approach. Test the bridge-balance condition; a balanced bridge carries no current in the bridging resistor, collapsing the network to two series arms in parallel plus the series $3\,\text{k}\Omega$.

Check bridge balance. A Wheatstone bridge is balanced when the ratio of the upper to lower arm is the same on both sides:$$\frac{R_{ua}}{R_{la}}=\frac{6\,\text{k}}{2\,\text{k}}=3,\qquad \frac{R_{ub}}{R_{lb}}=\frac{6\,\text{k}}{2\,\text{k}}=3.$$ The ratios are equal, so nodes $a$ and $b$ sit at the same potential.
Remove the bridging resistor. With $V_a=V_b$, the $6\,\text{k}\Omega$ between them carries zero current and can be deleted without changing anything:$$\boxed{\,I_{6\text{k, bridge}}=0\,}$$
Reduce the two series arms. Each side is now a simple series pair from the top node to the bottom node:$$R_{\text{left}}=6\,\text{k}+2\,\text{k}=8\,\text{k}\Omega,\qquad R_{\text{right}}=8\,\text{k}\Omega.$$ In parallel:$$R_{ab}=\frac{8\times 8}{8+8}\,\text{k}\Omega=4\,\text{k}\Omega.$$
Add the series resistor and apply Ohm’s law. The total resistance seen by the source is$$R_T=3\,\text{k}+4\,\text{k}=7\,\text{k}\Omega,$$ so$$I=\frac{V}{R_T}=\frac{50}{7000}=\boxed{\,7.14\ \text{mA}\,}$$

Quantity	Value
Bridging-resistor current	$0\text{ A}$ (balanced)
Equivalent bridge resistance $R_{ab}$	$4\,\text{k}\Omega$
Total resistance $R_T$	$7\,\text{k}\Omega$
Source current $I$	$7.14\text{ mA}$

Concept & Theory

Topic: Balanced Wheatstone bridge and series–parallel reduction.

Key relations: Balance condition $R_1/R_3=R_2/R_4$; parallel combination $R_\parallel=R_aR_b/(R_a+R_b)$; 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 bridge no matter what its resistance is. The bridge decouples into two independent series legs, which then combine in parallel. Recognising the balance is the graded insight; grinding a full node analysis wastes exam time.

Common pitfall: Assuming the centre resistor always matters. Always test balance first — if the ratios are equal the middle branch is irrelevant; if not, use a delta–wye transform on one triangle (the paper supplies the conversion formulae for exactly this case).

Source: Alexander & Sadiku, Fundamentals of Electric Circuits, 7th ed., §2.5–2.7 (Wheatstone bridge, wye–delta).

Question 1 of 6: Bridge network — source current

Question 1: Bridge network — source current [20]