Bridge network — equivalent resistance, current, power: 23-Mechatronics-A2 Circuits and Electronics, December 2018, Q1

Question 1: Bridge network — equivalent resistance, current, 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.

Element	Value
Source $V_{dc}$	100 V, A(+) to B(−)
Series resistor (carries I)	5 Ω
Bridge upper arms (T→L, T→R)	10 Ω each
Bridge middle arm (L→R)	10 Ω
Bridge lower arms (L→B, R→B)	15 Ω each
Direct outer branch (T→B)	12 Ω

Find. $R_{AB}$, the current I after the 5 Ω resistor, and the power dissipated in the 12 Ω resistor.

Figure-1: 100 V source, 5 Ω series resistor (current I), feeding a balanced Wheatstone bridge (10/10/10/15/15 Ω) in parallel with a direct 12 Ω branch from the same top node T to bottom node B.

Approach. Recognize the diamond is a balanced Wheatstone bridge (equal arm ratio 10:15 on both sides), collapse it, combine with the parallel 12 Ω branch, then add the series 5 Ω.

Test the bridge balance. Ratio T→L→B is $10\,\Omega$ then $15\,\Omega$; ratio T→R→B is the same $10\,\Omega$ then $15\,\Omega$. Since $\dfrac{10}{15}=\dfrac{10}{15}$, nodes L and R sit at the same potential, so no current flows in the middle 10 Ω bridge resistor — it can be removed without changing $R_{AB}$.
Collapse the bridge. With the bridge resistor removed, T–L–B ($10+15=25\,\Omega$) is in parallel with T–R–B ($10+15=25\,\Omega$): $$R_{\text{bridge}}=\frac{25\times25}{25+25}=\boxed{12.5\ \Omega}.$$
Combine with the direct 12 Ω branch. This 12.5 Ω sits in parallel with the outer 12 Ω branch, both from T to B: $$R_{TB}=\frac{12.5\times12}{12.5+12}=6.122\ \Omega.$$
Add the series 5 Ω. $$R_{AB}=5+R_{TB}=5+6.122=\boxed{11.12\ \Omega}.$$
Current I. I is the total source current (it flows through the 5 Ω before splitting at T): $$I=\frac{V_{dc}}{R_{AB}}=\frac{100}{11.12}=\boxed{8.99\ \text{A}}.$$
Power in the 12 Ω. The voltage across the T–B parallel combination is $V_{TB}=I\cdot R_{TB}=8.99\times6.122=55.05\ \text{V}$, shared by both parallel branches. The 12 Ω branch current is $$I_{12}=\frac{V_{TB}}{12}=\frac{55.05}{12}=4.59\ \text{A},$$ so $$P_{12\Omega}=I_{12}^2\times12=\boxed{252.5\ \text{W}}.$$

Quantity	Result
Equivalent resistance	$\boxed{R_{AB}=11.12\ \Omega}$
Source/series current	$\boxed{I=8.99\ \text{A}}$
Power in 12 Ω	$\boxed{P_{12\Omega}=252.5\ \text{W}}$

Concept & Theory

Topic. Balanced Wheatstone-bridge reduction combined with series–parallel resistance combination.

Key relations. Balance condition $R_1/R_2=R_3/R_4$ (equal arm ratios) $\Rightarrow$ zero bridge current; series $R=\sum R_i$; parallel $R=\big(\sum 1/R_i\big)^{-1}$; $P=I^2R$.

Why this works. A bridge is balanced exactly when the two voltage-divider ratios formed by its four outer arms are equal, which forces nodes L and R to the same potential regardless of the source value. With no potential difference across it, the middle (bridge) resistor carries no current and can be deleted — a purely topological simplification, not an approximation.

Common pitfall. Assuming a bridge must be redrawn with a delta–star transform every time; checking balance first often removes an element entirely and avoids the transform. Also, forgetting that I in part (b) is the total current (before the bridge splits it), not a branch current.

Source. Sadiku, Fundamentals of Electric Circuits §2.7 (series–parallel), §19 companion notes on bridge circuits.

Question 1 of 11: Bridge network — equivalent resistance, current, power

Question 1: Bridge network — equivalent resistance, current, power [10 + 5 + 5]