Question

a hobbyist building a radio needs a 150ω resistor in her circuit but has only a 220ω, a 79ω, and a 92ω resistor available. how can she connect these resistors to produce the desired resistance?

a. 268ω
b. 418ω
c. 591ω
d. 632ω
e. 150ω

Explanation:

Step1: Recall resistor - combination formulas

For resistors in series, $R_{total}=R_1 + R_2+R_3+\cdots$. For resistors in parallel, $\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots$.

Step2: Try parallel - combination of the given resistors

Let's try the parallel - combination of the three given resistors $R_1 = 220\Omega$, $R_2=79\Omega$, and $R_3 = 92\Omega$. Using the parallel - resistor formula $\frac{1}{R_{total}}=\frac{1}{220}+\frac{1}{79}+\frac{1}{92}$.
First, find a common denominator. The common denominator of 220, 79, and 92 is $220\times79\times92=220\times7268 = 1598960$.
$\frac{1}{R_{total}}=\frac{79\times92+220\times92 + 220\times79}{1598960}=\frac{7268+20240+17380}{1598960}=\frac{44888}{1598960}$.
Then $R_{total}=\frac{1598960}{44888}\approx35.62\Omega$. This is not correct.

Step3: Try a combination of series and parallel

Let's try combining the $79\Omega$ and $92\Omega$ resistors in series first. $R_{s1}=79 + 92=171\Omega$.
Then combine this series combination $R_{s1}$ with the $220\Omega$ resistor in parallel. Using the parallel - resistor formula $\frac{1}{R_{total}}=\frac{1}{220}+\frac{1}{171}$.
The common denominator is $220\times171 = 37620$.
$\frac{1}{R_{total}}=\frac{171+220}{37620}=\frac{391}{37620}$.
$R_{total}=\frac{37620}{391}\approx96.21\Omega$. This is not correct.

Step4: Another combination

Let's combine the $220\Omega$ and $79\Omega$ resistors in parallel. $\frac{1}{R_{p1}}=\frac{1}{220}+\frac{1}{79}=\frac{79 + 220}{220\times79}=\frac{299}{17380}$.
$R_{p1}=\frac{17380}{299}\approx58.13\Omega$.
Then combine this parallel combination $R_{p1}$ with the $92\Omega$ resistor in series. $R_{total}=58.13+92 = 150.13\approx150\Omega$.

Answer:

e. $150\Omega$