Question

Suppose that you have a 480 resistance, a 860 resistance, and a 1.20 k resistor.
(a) What is the maximum resistance you can obtain by combining these?
answer in k
(b) What is the minimum resistance you can obtain by combining these?
(answer in horseshoe looking symbol)

Answer 1

Concepts and reason

The main concept used to solve this problem is resistance. Initially, use the expression of equivalent resistance of a series combination to find the maximum resistance. Then, use the expression of equivalent resistance of a parallel combination to find the minimum resistance.

Fundamentals

The equivalent resistance of three resistances in series is as follows:

\(R_{\mathrm{s}}=R_{1}+R_{2}+R_{3}\)

Here, \(R_{1}, R_{2},\) and \(R_{3}\) are three resistances connected in series. The equivalent resistance of three resistances in parallel is as follows:

\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)

Here, \(R_{1}, R_{2},\) and \(R_{3}\) are three resistances connected in parallel.

 

(a) The equivalent resistance of three resistances in series is as follows:

\(R_{\mathrm{s}}=R_{1}+R_{2}+R_{3}\)

Substitute \(480 \Omega\) for \(R_{1}, 860 \Omega\) for \(R_{2},\) and \(1.20 \mathrm{k} \Omega\) for \(R_{3}\) in the above expression.

$$ \begin{array}{c} R_{\mathrm{s}}=480 \Omega+860 \Omega+(1.20 \mathrm{k} \Omega)\left(\frac{1000 \Omega}{1 \mathrm{k} \Omega}\right) \\ =(2540 \Omega)\left(\frac{1 \mathrm{k} \Omega}{1000 \Omega}\right) \\ =2.50 \mathrm{k} \Omega \end{array} $$

Part a

The maximum possible resistance is \(2.50 \mathrm{k} \Omega\).

The maximum resistance can be obtained when all the resistors will add up. This is only possible when the three resistors are connected in series.

 

(b) The equivalent resistance of three resistances in parallel is as follows:

\(\frac{1}{R_{\mathrm{p}}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)

Substitute \(480 \Omega\) for \(R_{1}, 860 \Omega\) for \(R_{2},\) and \(1.20 \mathrm{k} \Omega\) for \(R_{3}\) in the above expression.

$$ \begin{array}{c} \frac{1}{R_{\mathrm{p}}}=\frac{1}{480 \Omega}+\frac{1}{860 \Omega}+\frac{1}{1.20 \mathrm{k} \Omega}\left(\frac{1 \mathrm{k} \Omega}{1000 \Omega}\right) \\ \frac{1}{R_{\mathrm{p}}}=4.079457 \Omega^{-1} \\ R_{\mathrm{p}}=245.13 \Omega \\ =245 \Omega \end{array} $$

Part b

The minimum possible resistance is \(245 \Omega\).

The minimum resistance can be obtained when all the resistors are connected in a parallel combination. Any other combination will give a higher value of equivalent resistance.