Question

You have six 1.5 kω resistors.

Answer 1

Answer:

Have three pairs of 1.5kΩ connected in series and then connect all three in parallel.

Explanation:

This problem is kind of hard to explain and is based off my experience and intuition. I'll show how I got the answer and then explain afterwards.

If you have three pairs of 1.5kΩ in series, you get 3 3.0kΩ resistor equivalent (resistors in series). Putting them in parallel, you use get a resistance of 1 kΩ as shown below:

$R_{eq} = (\frac{1}{R_{1}} + \frac{1}{R_{2}} +...+\frac{1}{R_{n}} )^{-1} \\R_{eq} = (\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} )^{-1}\\R_{eq} = (\frac{1}{3} + \frac{1}{3} + \frac{1}{3} )^{-1}\\ R_{eq} = 1$

How I got here takes some math sense. Since 1.5kΩ is greater than 1.0kΩ, you know there is a parallel combination. Next I considered what are the possible combinations. Since two 1.5kΩ resistors yielded 3.0kΩ, and you have 3 pairs of 3.0kΩ resistor, the sum of parallel would be 1.0kΩ