DC bridge network — equivalent resistance, current and power: 22-Elec-A1 Circuits, December 2014, Q1

Question 1: DC bridge network — equivalent resistance, current and power [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.

Given. A $100\text{ V}$ DC source drives a $5\,\Omega$ series resistor (carrying $I$) into a resistive bridge whose top vertex $P$ and bottom vertex $Q$ are also bridged by a $12\,\Omega$ resistor:

Element	Value	Connection
Source $V_{dc}$	$100$ V	A(+) – B($-$)
Series $R$	$5\,\Omega$	A $\to$ P (carries $I$)
Upper arms	$10\,\Omega,\;10\,\Omega$	P–L and P–R
Bridge arm	$10\,\Omega$	L–R (galvanometer branch)
Lower arms	$15\,\Omega,\;15\,\Omega$	L–Q and R–Q
Load $R$	$12\,\Omega$	P $\to$ B (parallel with bridge)

Find. $R_{AB}$; the source current $I$; and the power in the $12\,\Omega$ resistor.

[Figure not reproduced: Figure-1 (redrawn). The diamond bridge P–L–R–Q sits between the top rail (P) and the bottom rail (Q = B), and the $12\,\Omega$ resistor spans the same two rails, so it is in parallel with the bridge; the $5\,\Omega$ is in series ahead of them. See the official exam paper.]

Approach. Test the bridge for balance (if balanced the centre arm carries no current and drops out), reduce the bridge to a single resistance $R_{PQ}$, combine it in parallel with the $12\,\Omega$, add the series $5\,\Omega$, then apply Ohm’s law and $P=V^2/R$.

Check bridge balance. A Wheatstone bridge is balanced when the ratios of the two arms on each side are equal: $$\frac{R_{PL}}{R_{LQ}}=\frac{10}{15}=\frac{R_{PR}}{R_{RQ}}=\frac{10}{15}.$$ The ratios match, so nodes L and R sit at the same potential and the central $10\,\Omega$ arm carries $\boxed{0\ \text{A}}$ — it may be removed.
Reduce the bridge $R_{PQ}$. With the centre arm gone, each side is a simple series pair, and the two sides are in parallel between P and Q: $$R_{PQ}=(10+15)\,\|\,(10+15)=25\,\|\,25=\boxed{12.5\ \Omega}.$$
Bridge in parallel with the $12\,\Omega$. Both span P–B: $$R_{PB}=R_{PQ}\,\|\,12=\frac{12.5\times12}{12.5+12}=\frac{150}{24.5}=6.122\ \Omega.$$
Equivalent resistance (a). Add the series $5\,\Omega$: $$R_{AB}=5+R_{PB}=5+6.122=\boxed{11.12\ \Omega}.$$
Source current (b). Ohm’s law across A–B: $$I=\frac{V_{dc}}{R_{AB}}=\frac{100}{11.12}=\boxed{8.99\ \text{A}}.$$
Power in the $12\,\Omega$ (c). The voltage across the parallel block equals the voltage across the $12\,\Omega$: $V_{PB}=I\,R_{PB}=8.99\times6.122=55.05\ \text{V}$. Hence $$P_{12}=\frac{V_{PB}^{2}}{12}=\frac{55.05^{2}}{12}=\boxed{252.5\ \text{W}}.$$

Quantity	Result
$R_{PQ}$ (bridge)	$12.5\ \Omega$ (balanced)
$R_{AB}$	$\boxed{11.12\ \Omega}$
$I$	$\boxed{8.99\ \text{A}}$
$P_{12\Omega}$	$\boxed{252.5\ \text{W}}$

Concept & Theory

Topic. Series–parallel reduction and the balanced Wheatstone bridge.

Key relations. Bridge balance $R_1/R_2=R_3/R_4$; series $R=\sum R_i$; parallel $R=(\tfrac1{R_a}+\tfrac1{R_b})^{-1}$; Ohm $I=V/R$; power $P=V^2/R=I^2R$.

Why this works. When the bridge is balanced the two mid-nodes are equipotential, so no current crosses the bridge arm regardless of its value; the network then collapses to two series legs in parallel. Recognising the balance up front avoids a four-node system. The load resistor shares the same node pair as the bridge, so it parallels the reduced bridge rather than adding in series — the single most common misread of this figure.

Common pitfall. Treating the $12\,\Omega$ as being in series with the bridge (it is not) gives $R_{AB}=29\,\Omega$ and badly wrong current and power. Also, once balance is established, do not assign the centre arm any current.

Source. Alexander & Sadiku, Fundamentals of Electric Circuits, §2.5–2.7 (series/parallel, Wheatstone bridge).

Question 1 of 6: DC bridge network — equivalent resistance, current and power

Question 1: DC bridge network — equivalent resistance, current and power [10 + 5 + 5]