Equivalent resistance in a triangular resistor network

Steps:

• Network: R between X-Y, 5 Ω between Y-Z, R between X-Z.
• Y-Z resistance: 5 Ω parallel with 2R equals 2.5 Ω, so \frac{10R}{5+2R} = 2.5.
• Solve: 10R = 12.5 + 5R, so 5R = 12.5, R = 2.5 Ω.
• X-Y resistance: 2.5 Ω parallel with (2.5 + 5) = 7.5 Ω, so \frac{2.5 \times 7.5}{10} = 1.875 Ω \approx 1.9 Ω.

Why C is correct:

• Parallel combination formula gives exactly 1.875 Ω, matching 1.9 Ω option.

Why the others are wrong:

• A. 0.3 Ω: Underestimates by ignoring proper parallel paths.
• B. 0.5 Ω: Results from incorrect series assumption for Y-Z.
• D. 3 Ω: Approximates total network resistance, not X-Y specifically.

Final answer: C