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Chapter 25: Problem 19

A \(47-\mathrm{k} \Omega\) resistor and a \(39-\mathrm{k} \Omega\) resistor are in parallel, and the pair is in series with a \(22-\mathrm{k} \Omega\) resistor. What's the resistance of the combination?

Short Answer

Expert verified
The resistance of the combination is calculated by first determining the equivalent resistance of the parallel resistors, then adding this to the resistance of the series resistor.

Step by step solution

01

Resistance of Parallel Resistors

Firstly, calculate the resistance of the parallel combination of \(47-k \Omega\) and \(39-k \Omega\) resistors. According to the formula for resistors in parallel, \(\frac{1}{R_{parallel}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\). Then calculate \(R_{parallel}\) by taking the reciprocal of the result.
02

Resistance of Resistors in Series

Now, calculate the resistance of the pair of parallel resistors being in series with the \(22-k \Omega\) resistor. The resistance for resistors in series is calculated as \(R_{total} = R_{1} + R_{2}\). It means, the total resistance would be \(R_{total} = R_{parallel} + 22k\Omega\).
03

Evaluate the Total Resistance

Having calculated the resistances in both parallel and series arrangements, find the total resistance which is the sum of \(R_{parallel}\) and \(22-k \Omega\) resistor. This will be the final answer.

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Most popular questions from this chapter

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