A 35 kΩ is connected in series with a resistance of 40 kΩ. What resistance R must be connected in parallel…

Analyze the question: Two resistors (35 kΩ and 40 kΩ) are in series. A third resistor R is placed in parallel with that combination. The total equivalent resistance must be 25 kΩ. We need R.

Concept:

• Series: R_series = R₁ + R₂
• Parallel: 1/R_eq = 1/R_series + 1/R

Solution:

Step 1: Series combination R_series = 35 + 40 = 75 kΩ

Step 2: Parallel with R 1/25 = 1/75 + 1/R

Step 3: Solve for R 1/R = 1/25 − 1/75 = 3/75 − 1/75 = 2/75 R = 75/2 = 37.5 kΩ

Check: (75 × 37.5)/(75 + 37.5) = 2812.5/112.5 = 25 kΩ ✓

Answer: R = 37.5 kΩ, which is option B.

Tip: For a parallel resistor needed to reach a target, use R = (R_s × R_eq)/(R_s − R_eq). Here: (75 × 25)/(75 − 25) = 1875/50 = 37.5 kΩ. Also remember that the parallel equivalent is always smaller than the smallest branch, so the 25 kΩ target must be below 75 kΩ, which it is.