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

Question 1: 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.

[Figure not reproduced: Figure‑1 — redrawn from the printed figure. A balanced 10/10/15/15 Wheatstone bridge (centre 10Ω) with the 12Ω across the same top–bottom rails, so the 12Ω parallels the bridge. See the official exam paper.]

Given. A $V_{dc}=100\text{ V}$ source drives terminals A–B. After a $5\,\Omega$ input resistor (carrying the marked current $I$) the network is a Wheatstone bridge whose four arms are $10\,\Omega,10\,\Omega$ (top) and $15\,\Omega,15\,\Omega$ (bottom) with a $10\,\Omega$ bridging resistor across the two mid-nodes. The $12\,\Omega$ resistor connects the bridge’s top drive node to the bottom rail — i.e. it sits directly across the same two nodes as the whole bridge.

Find. (a) $R_{AB}$; (b) $I$; (c) $P_{12\Omega}$.

Approach. Test the bridge for balance; a balanced bridge carries no current in the centre arm, so it collapses to two series legs in parallel, after which the $12\,\Omega$ (across the same nodes) and the $5\,\Omega$ input reduce by inspection, and Ohm’s law gives the current and power.

1. Check the bridge balance. With the top drive node $T$ and bottom drive node $G$, the balance condition for the centre arm is $\dfrac{R_{T\text{-}L}}{R_{L\text{-}G}}=\dfrac{R_{T\text{-}R}}{R_{R\text{-}G}}$. Here $\dfrac{10}{15}=\dfrac{10}{15}$, so the bridge is balanced and the centre $10\,\Omega$ carries zero current — it can be removed.
2. Reduce the balanced bridge. Each side is then a simple series pair, $10+15=25\,\Omega$, and the two sides are in parallel between $T$ and $G$: $$R_{br}=25\,\|\,25=\frac{25\cdot25}{50}=\boxed{12.5\,\Omega.}$$
3. Combine with the $12\,\Omega$. The $12\,\Omega$ spans the same nodes $T\!-\!G$, so it parallels the bridge: $$R_{TG}=12.5\,\|\,12=\frac{12.5\cdot12}{24.5}=6.122\,\Omega.$$
4. Equivalent resistance at A–B. Add the $5\,\Omega$ input in series: $$\boxed{R_{AB}=5+6.122=11.12\,\Omega.}$$
5. Current $I$. $I$ is the current the source pushes through the $5\,\Omega$, i.e. the total current: $$\boxed{I=\frac{V_{dc}}{R_{AB}}=\frac{100}{11.12}=8.99\text{ A}.}$$
6. Power in the $12\,\Omega$. The voltage across the parallel block is $V_{TG}=I\,R_{TG}=8.99(6.122)=55.05\text{ V}$, so $$\boxed{P_{12\Omega}=\frac{V_{TG}^{\,2}}{12}=\frac{55.05^{2}}{12}=252.5\text{ W}.}$$