How do you solve this? Your lab contains a large number of resistors in the foll
ID: 1922498 • Letter: H
Question
How do you solve this?
Your lab contains a large number of resistors in the following standard value 1. 8 Ohm , 20 Ohm , 300 Ohm , 24 Ohm , 56 Ohm , Your circuit design requires the following resistances. Find the most efficient (i. e. fewest total resistors) combination of resistors to achieve these resistances 5 Ohm , 311. 8 Ohm , 40 k Ohm , 52. 32 k Ohm In engineering design problems, we are frequently given specifications that have some leeway For example, in part 11. 1a. we could be asked to create a resistance of 5 Ohm plusminus 10%, that is, 4. 5 Ohm R 5. 5 Ohm . Can you find a more efficient (fewer resistors) combination of resistors for each of the resistances of 11. 1 given a plusminus tolerance?Explanation / Answer
It will help to realize that putting two resistors in parallel gives you an equivalent resistance that is half of the resistance of each resistor by itself.
11.1
a) Put 4 20 resistors in parallel to get 5
b) First, put 2 20 resistors in parallel to get a 10 resistor. Put that in series with a 300 resistor and a 1.8 resistor to get 311.8.
c) Make two 80k-equivalent resistors by putting a 24k and 56k resistor in series. Put these two 80k-equivalent resistors in parallel to get 40k.
d) First, realize that 52.32k is just 52,320. Let's think of this as 52k + 320. To get the 52k, first put two 56k resistors in parallel to obtain 28k. Then put a 24k resistor in series to obtain 52k. Put a 300 resistor and 20 resistor in series with that to get 52.32k.
11.2
a) 5.4 is within the range. So, put three 1.8 resistors in series.
b) Our range here is 280.62 R 342.98. Just one 300 resistor is in this range.
c) Our range here is 36k R 44k. Put two 24k resistors in parallel to get a 12k-equivalent resistor. Put this in series with another 24k resistor to get 36k.
d) Our range here is 47.088k R 57.552k. One 56k resistor is in this range.
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