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Chapter 25: Problem 20

What resistance should you place in parallel with a \(56-\mathrm{k} \Omega\) resistor to make an equivalent resistance of \(45 \mathrm{k} \Omega ?\)

Short Answer

Expert verified
The resistance required for the specified parallel arrangement is a \(252 k\Omega\) resistor.

Step by step solution

01

Understand the given values

From the problem, the value of the first resistor \(R1 = 56k \Omega\) and the equivalent resistance \(R_{eq} = 45k \Omega\). The value of the secondary resistor required \(R2\) is to be found.
02

Rearrange the formula to find unknown resistance \(R2\)

Rearrange the formula \(\frac{1}{R_{eq}} = \frac{1}{R1} + \frac{1}{R2}\) for \(R2\). We get, \(R2 = \frac{1}{(\frac{1}{R_{eq}} - \frac{1}{R1})}\). This will help us find the value of unknown resistance.
03

Substitute the known values into the derived formula

Plugging in the known values \(R1 = 56k \Omega\) and \(R_{eq} = 45k \Omega\) into the formula from Step 2, we get \(R2 = \frac{1}{(\frac{1}{45k} - \frac{1}{56k})}\)
04

Solve the equation

By evaluating the equation from Step 3, we get \(R2 \approx 252k \Omega\). Therefore, a resistor with a resistance of approximately \(252k \Omega\) should be placed parallel to the \(56k \Omega\) resistor to establish an equivalent resistance of \(45k \Omega\) as stated in the given problem.

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