A 35 kΩ is connected in series with a resis..

Question 1 Report

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

Answer Details

For the combination in series;

⇒R1 = 35kΩ + 40kΩ = 75kΩ

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

= $\frac{1}{R_{e} q}$ = $\frac{1}{R}$ + $\frac{1}{R}$

= $\frac{1}{25}$ = $\frac{1}{R}$ + $\frac{1}{75}$

= $\frac{1}{25}$ - $\frac{1}{75}$ + $\frac{1}{R}$

= $\frac{3 - 1}{75}$ = $\frac{1}{R}$

= $\frac{2}{75}$ = $\frac{1}{R}$

= $\frac{75}{2}$ = R

; R = 37.5k Ω

Read lesson note on Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

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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 resis...