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?
A
40 kΩ
B
37.5 kΩ
C
45.5 kΩ
D
30 kΩ
Verified Correct Answer: B.
Physical Principles & Numerical Solution
For the combination in series;
⇒R1 = 35k\Omega + 40k\Omega = 75k\Omega
R is combined with 75k\Omega in parallel to give 25k\Omega
= \(\frac{1}{R_eq}\) = \(\frac{1}{R}\) + \(\frac{1}{R}\)
= \(\frac{1}{25}\) = \(\frac{1}{R}\) + \(\frac{1}{75}\)
= \(\frac{1}{25}\) - \(\frac{1}{75}\) + \(\frac{1}{R}\)
= \(\frac{3-1}{75}\) = \(\frac{1}{R}\)
= \(\frac{2}{75}\) = \(\frac{1}{R}\)
= \(\frac{75}{2}\) = R
; R = 37.5k \Omega
Source: WAEC / JAMB / NECO Classroom Archive (Record #68011)