Current and voltage across the bridge resistor R₅: 22-Elec-A5 Electronics, May 2018, Q1

Question 1: Current and voltage across the bridge resistor R₅

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. See the official exam paper or the cited reference text.]

A, right leg R₃–R₄ taps v_B, and R₅ bridges v_A–v_B." style="max-width:460px;width:100%;height:auto">

Question 1 — resistive bridge: +10 V at the top node, bottom node grounded; left leg R₁–R₂ taps v_A, right leg R₃–R₄ taps v_B, and R₅ bridges v_A–v_B.

Given. A four-arm resistive (Wheatstone) bridge driven by a 10 V source at the top node with the bottom node grounded; R₅ bridges the two tap nodes v_A and v_B.

Given data
Quantity	Value
Source	+10 V (top), 0 V (bottom)
R₁, R₂	1 kΩ, 1.2 kΩ
R₃, R₄	9.1 kΩ, 11 kΩ
R₅	2 kΩ

Find. The current $i_5$ through R₅ (arrow directed v_A→v_B) and the bridge voltage $v_{AB}=v_A-v_B$.

Approach. Write one KCL (node-voltage) equation at each tap node with the top rail fixed at 10 V and the bottom at 0 V, solve the 2×2 system, then read $i_5$ and $v_{AB}$ directly.

KCL at node A. Sum the currents leaving v_A through R₁ (to +10 V), R₂ (to ground) and R₅ (to v_B):$$\frac{v_A-10}{R_1}+\frac{v_A}{R_2}+\frac{v_A-v_B}{R_5}=0.$$
KCL at node B. Likewise at v_B:$$\frac{v_B-10}{R_3}+\frac{v_B}{R_4}+\frac{v_B-v_A}{R_5}=0.$$
Solve the pair. Substituting the resistor values and solving the two linear equations gives $v_A=5.456\ \text{V}$ and $v_B=5.461\ \text{V}$. The two open-circuit divider ratios are almost equal — $R_2/(R_1+R_2)=0.5455$ versus $R_4/(R_3+R_4)=0.5473$ — so the bridge is only slightly unbalanced and the tap nodes sit within a few millivolts of each other.
Bridge voltage and current. $v_{AB}=v_A-v_B$, and $i_5=v_{AB}/R_5$:$$v_{AB}=5.456-5.461=\boxed{-4.81\ \text{mV}},\qquad i_5=\frac{-4.81\ \text{mV}}{2\ \text{k}\Omega}=\boxed{-2.40\ \mu\text{A}}.$$ The negative sign means the actual current flows from v_B to v_A (opposite to the reference arrow), because v_B is the marginally higher node.

Question 1 — results
Quantity	Value
$v_A$, $v_B$	5.456 V, 5.461 V
Bridge voltage $v_{AB}$	−4.81 mV
Bridge current $i_5$	−2.40 µA (flows B→A)

Topic. Balanced/unbalanced resistive (Wheatstone) bridge solved by node-voltage analysis.

Key relations. Balance (zero bridge current) requires equal divider ratios, $R_1/R_2=R_3/R_4$. Here $1/1.2=0.833$ versus $9.1/11=0.827$, a 0.7 % mismatch, so a small residual current flows through the bridge arm.

Why this works. With both rail voltages fixed, only the two tap-node voltages are unknown; two KCL equations determine them exactly, and the bridge branch current follows from Ohm’s law across R₅.

Common pitfall. Do not assume the bridge is balanced just because the ratios look close — the residual is small but real, and dropping R₅ (treating it as an ideal galvanometer) changes the tap voltages slightly. Keep R₅ in the node equations.

Source. Alexander & Sadiku, Fundamentals of Electric Circuits, 8th ed., Ch. 3 (nodal analysis; Wheatstone bridge).

Question 1 of 5: Current and voltage across the bridge resistor R₅

Question 1: Current and voltage across the bridge resistor R₅