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A resistor bridge has four nodes A, B, C, D. The resistors are: A-C =…
Question
A resistor bridge has four nodes A, B, C, D. The resistors are: A-C = 2 ohm, A-D = 4 ohm, C-B = 8 ohm, D-B = 6 ohm, and a bridging resistor C-D = 10 ohm. The bridge is unbalanced. Let the equivalent resistance between A and B be R = p/q ohm in lowest terms. Report p + q.
✓ Verified answer: 421checked by our engine — not a guess
Step-by-step solution
Drive 1 A into A, extract at B, set V_B = 0.
Write Kirchhoff current law at the inner nodes C and D (sum of currents leaving = 0) and at A (1 A enters).
With conductances 1/R per branch:
Node C: (V_C-V_A)/2 + V_C/8 + (V_C-V_D)/10 = 0.
Node D: (V_D-V_A)/4 + V_D/6 + (V_D-V_C)/10 = 0.
Node A: (V_A-V_C)/2 + (V_A-V_D)/4 = 1.
Solving the linear system gives V_A = 350/71 V. Since 1 A flows in, R_AB = V_A = 350/71 ohm, already in lowest terms. Hence p+q = 350+71 = 421.
Final answer421
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