A 35 kΩ is connected in series with a resistance of 40 kΩ. What resistance R must be connected in parallel with the combination so that the equivalent resistance is equal to 25 kΩ?
- 40 kΩ
-
37.5 kΩ
- 45.5 kΩ
- 30 kΩ
The correct answer is: B
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
There is an explanation video available .