NivaarExam PrepOfficial exam papers ↗

22-Elec-A5 Electronics · Undated paper

Question 1 of 5: Resistive bridge — current and voltage across R 5

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — Electronics, 16-Elec-A5. Closed-book; one approved Casio/Sharp calculator. Answer all five questions; each worth 20 marks. Op-amps assumed ideal with ±15 V supplies unless stated; ground and chassis common. Duration 3 hours.

Reference texts. A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. (diodes ch. 4; MOSFET & BJT amplifiers ch. 5–8; MOS inverters ch. 13–14) — primary; C. K. Alexander & M. Sadiku, Fundamentals of Electric Circuits, 7th ed. (Thévenin equivalents, ch. 4) for the resistive-bridge question.

The question pages identify this paper as the 16-Elec-A5 Electronics, May 2019 sitting.

Question 1: Resistive bridge — current and voltage across 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.

Given. A Wheatstone (diamond) bridge driven from a $+10\text{ V}$ rail to ground, with $R_5$ bridging the two mid-nodes $v_A$ and $v_B$.

Given data
$R_1$$1\text{ k}\Omega$$R_2$$1.2\text{ k}\Omega$
$R_3$$9.1\text{ k}\Omega$$R_4$$11\text{ k}\Omega$
$R_5$$2\text{ k}\Omega$Source$+10\text{ V}$

Find. The current $i_5$ (defined $A\!\to\!B$) and the voltage $v_{AB}=v_A-v_B$.

+10 V R₁ v A R₂ R₃ v B R₄ R₅ i 5
Bridge network. Left arm $R_1/R_2$ sets $v_A$; right arm $R_3/R_4$ sets $v_B$; $R_5$ carries $i_5$ from $v_A$ to $v_B$.

Approach. Replace everything to the left of $R_5$ (the two dividers) by its Thévenin equivalent across $A\text{-}B$, then $R_5$ is a single series load.

  1. Open-circuit node voltages (remove $R_5$). Each arm is a plain divider from $10\text{ V}$: $v_A^{oc}=10\,\dfrac{R_2}{R_1+R_2}=10\cdot\dfrac{1.2}{2.2}=5.4545\text{ V}$, $v_B^{oc}=10\,\dfrac{R_4}{R_3+R_4}=10\cdot\dfrac{11}{20.1}=5.4726\text{ V}$.
  2. Thévenin voltage. $V_{th}=v_A^{oc}-v_B^{oc}=5.4545-5.4726=\boxed{-18.1\text{ mV}}$ — the bridge is nearly balanced ($R_1/R_2\approx R_3/R_4$), so the drive across $R_5$ is small.
  3. Thévenin resistance. Kill the source (short the $10\text{ V}$ rail); each arm collapses to its two resistors in parallel: $R_{th}=(R_1\Vert R_2)+(R_3\Vert R_4)=\dfrac{1\cdot1.2}{2.2}+\dfrac{9.1\cdot11}{20.1}=0.545+4.980=5.526\text{ k}\Omega$.
  4. Loaded current. $R_5$ sees $V_{th}$ through $R_{th}$: $i_5=\dfrac{V_{th}}{R_{th}+R_5}=\dfrac{-18.1\text{ mV}}{5.526+2\text{ k}\Omega}=\dfrac{-18.1\text{ mV}}{7.526\text{ k}\Omega}=\boxed{-2.40\ \mu\text{A}}$. The negative sign means the current actually flows $B\!\to\!A$.
  5. Voltage across $R_5$. $v_{AB}=i_5\,R_5=(-2.40\ \mu\text{A})(2\text{ k}\Omega)=\boxed{-4.81\text{ mV}}$.
Question 1 results
$v_A^{oc}$$v_B^{oc}$$V_{th}$$R_{th}$$i_5$$v_{AB}$
$5.4545\text{ V}$$5.4726\text{ V}$$-18.1\text{ mV}$$5.53\text{ k}\Omega$$-2.40\ \mu\text{A}$$-4.81\text{ mV}$
← Paper overview