Resistor Bridge — Current and Voltage in R 5: 22-Elec-A5 Electronics, December 2014, Q2

Question 2: Resistor Bridge — Current and Voltage in R₅ (20 marks)

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.

Given data
Source	+10 V (top node), bottom node grounded
Left arm	$R_1=1\ \text{k}\Omega$ (top), $R_2=1.2\ \text{k}\Omega$ (bottom); junction $v_A$
Right arm	$R_3=9.1\ \text{k}\Omega$ (top), $R_4=11\ \text{k}\Omega$ (bottom); junction $v_B$
Bridge	$R_5=2\ \text{k}\Omega$ from $v_A$ to $v_B$, current $i_5$ defined $A\to B$

Find. The current $i_5$ through $R_5$ and the voltage $V_{R5}=v_A-v_B$ across it.

Q2: resistor bridge fed by +10 V. R5 bridges node v_A (R1-R2 junction) to node v_B (R3-R4 junction); find i5 and V_R5.

Approach. Remove $R_5$ and find the Thévenin equivalent the bridge presents to it (open-circuit voltage between $v_A$ and $v_B$, and the resistance looking back), then reconnect $R_5$ as a single series load.

Open-circuit node voltages (R5 removed). Each arm is a simple divider from 10 V: $$v_{A,oc}=10\cdot\frac{R_2}{R_1+R_2}=10\cdot\frac{1.2}{2.2}=5.4545\text{ V},\qquad v_{B,oc}=10\cdot\frac{R_4}{R_3+R_4}=10\cdot\frac{11}{20.1}=5.4726\text{ V}.$$ The Thévenin voltage is $V_{th}=v_{A,oc}-v_{B,oc}=-0.0181\text{ V}$ — the bridge is very nearly balanced ($R_1/R_2=0.833$ versus $R_3/R_4=0.827$).
Thévenin resistance. Deactivate the source (10 V node → ground); looking back into the $v_A,v_B$ terminals each arm becomes its two resistors in parallel: $$R_{th}=(R_1\!\parallel\!R_2)+(R_3\!\parallel\!R_4)=0.5455+4.9801=5.5256\ \text{k}\Omega.$$
Reconnect R5 and solve. $R_5$ sees $V_{th}$ in series with $R_{th}$: $$\boxed{\,i_5=\frac{V_{th}}{R_{th}+R_5}=\frac{-0.0181}{5.5256+2}\text{ (k}\Omega)=-2.40\ \mu\text{A}\,}.$$ The negative sign means the actual current flows from $B$ to $A$ (opposite the drawn arrow).
Voltage across R5. $$\boxed{\,V_{R5}=v_A-v_B=i_5R_5=(-2.40\ \mu\text{A})(2\ \text{k}\Omega)=-4.81\ \text{mV}\,}.$$ A full two-node solve with $R_5$ in place gives $v_A=5.4559$ V, $v_B=5.4607$ V and reproduces the same $i_5=-2.40\ \mu\text{A}$, confirming the Thévenin result.

Physically the bridge is close to null: the tiny imbalance drives only a few microamps through the detector arm, which is the regime an instrumentation bridge is designed to operate in.

Q2 results
Quantity	Value
$v_{A,oc}$	5.4545 V
$v_{B,oc}$	5.4726 V
$R_{th}$	5.526 kΩ
$i_5$	−2.40 µA (flows $B\!\to\!A$)
$V_{R5}$	−4.81 mV

Topic: Wheatstone bridge / Thévenin reduction.

Key relations: voltage divider $v=V_s R_{low}/(R_{high}+R_{low})$; Thévenin $V_{th}=v_{A,oc}-v_{B,oc}$, $R_{th}=(R_1\parallel R_2)+(R_3\parallel R_4)$; bridge balance when $R_1/R_2=R_3/R_4$.

Why this works: $R_5$ couples the two arms, so neither node is a clean divider once it is connected. Replacing the rest of the network by its Thévenin equivalent turns the two-node problem into a single series loop that is solved by inspection, and the sign of $i_5$ falls out of the sign of $V_{th}$.

Common pitfall: Computing each node as an isolated divider with $R_5$ still attached (it is not — $R_5$ loads both nodes), or dropping the sign and mis-stating the current direction.

Source: Sadiku, Fundamentals of Electric Circuits, Thévenin’s theorem & the Wheatstone bridge (§4).

Question 2 of 5: Resistor Bridge — Current and Voltage in R 5

Question 2: Resistor Bridge — Current and Voltage in R5 (20 marks)

Question 2: Resistor Bridge — Current and Voltage in R₅ (20 marks)