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A 35 kΩ is connected in series with a resistance of 40 kΩ. What resistance ...
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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
Show Answer
Show Explanation
Correct Answer: 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
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