A 35 kΩ is connected in series with a resistance of 40 kΩ. What resistance...

Explanation

For the combination in series;

⇒R1 = 35kΩ + 40kΩ = 75kΩ

R is combined with 75kΩ in parallel to give 25kΩ

= \(\frac{1}{R_eq}\) = \(\frac{1}{R}\) + \(\frac{1}{R}\)

= \(\frac{1}{25}\) = \(\frac{1}{R}\) + \(\frac{1}{75}\)

= \(\frac{1}{25}\) - \(\frac{1}{75}\) + \(\frac{1}{R}\)

= \(\frac{3-1}{75}\) = \(\frac{1}{R}\)

= \(\frac{2}{75}\) = \(\frac{1}{R}\)

= \(\frac{75}{2}\) = R

; R = 37.5k Ω