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

Correct Option: Answer is B

To find the equivalent resistance of the combination, let’s denote the given resistances as follows:

R1=35kΩ (the first resistor)

R2=40kΩ (the second resistor)

The total resistance (Rtotal) of two resistors in series is the sum of their individual resistances:

Rtotal=R1+R2

Rtotal=35kΩ+40kΩ

Rtotal=75kΩ

Now, let R be the resistance connected in parallel with the combination. The formula for the total resistance of two resistors in parallel is given by:

Rtotal1=R31+R1

Given that Rtotal should be equal to 25 kΩ, we can write the equation as:

1/25 kΩ=1/75 kΩ+1/R

Now, solve for R:

1/R=1/25 kΩ−1/75 kΩ

1/R=3/75 kΩ−1/75 kΩ

1/R=2/75 kΩ

R=75 kΩ/2

R=37.5kΩ

Therefore, the resistance R that must be connected in parallel is:

B. 37.5 kΩ

## 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?

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