# getting answers to worksheet questions

Discussion in 'General Electronics Chat' started by ninjaman, May 20, 2016.

1. ### ninjaman Thread Starter Member

May 18, 2013
307
1
hello,
i am working through the voltage divider worksheet. i got to question 20 and couldn't figure out how to calculate it. i got most of the previous questions correct.
there are three pots in series that make up a voltage divider. the top and bottom pots are 5k with the middle pot 100k, the outputs of the 5k pots are connected to the inputs of the 100k pot. the setting for the pots is measured in percentages,

Note that the upper 5 kΩ potentiometer is set to its 20% position (m = 0.2), while the lower 5 kΩ potentiometer is set to its 90% position (m = 0.9), and the 100 kΩ potentiometer is set to its 40% position (m = 0.4).

so i figured that the top resistor was 4000 ohm before the 100k and 1000 ohm in parallel with the 100k. the bottom 5k has 500 ohms in parallel with the 100k and 4500 ohms to ground. the 1000 and 500 in parallel with 100k is 1477.8 ohms this is then in series with the 4500 ohms making 5977.8 ohms, this is R2. this is where i am getting stuck. i cant get the correct answer. i add the 5977 to the top 4000 and get 9977.8 which is almost the total resistance answer which is 9978 ohms. the voltage part is the problem. the 100k is in parallel with the 1500 ohms but does the 40% affect the total parallel value, or does it affect the 100k on its own.

question 21
Which voltage divider circuit will be least affected by the connection of identical loads? Explain your answer.

What advantage does the other voltage divider have over the circuit that is least affected by the connection of a load?
for this i thought that the current through the first divider would be higher and the load would not affect the circuit too much. but the answer states that the second divider wastes less energy? i am not sure why this is?

any help would be great!
thanks
simon

2. ### WBahn Moderator

Mar 31, 2012
23,208
6,997
The top pot is, indeed, split into 4 kΩ above the wiper and 1 kΩ below. But the 1 kΩ is NOT in parallel with the 100 kΩ pot. To be in parallel, the ends of the two components in question have to be at the same voltages. The 1000 Ω from the bottom of the top pot is in series with the 500 Ω from the top of the bottom pot. This is seen because whatever current flows in one HAS to flow in the other -- hence they are in series.

Replace the three pots with two appropriate resistors and give it a try.

There are several reasonable ways to tackle the analysis. One would be to first determine the total current in the circuit (i.e., what the source is putting out). Then use current division to determine how much goes through the 100 kΩ pot. Then find Vout by walking up the voltage drops from ground.

For the second problem, your reasoning is fine. The first divider has a higher current draw and the load will not affect it AS MUCH as the second one (you have no idea whether it will affect it "too much" because you have no idea what "too much" means for that load). The second part of the question asks what advantage does the OTHER circuit (the second circuit) have over the first. Well, you said that the first diver has a higher current draw, so the second circuit has a lower current draw, right? Isn't a circuit that draws less current (all else being equal) waste less energy?

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3. ### GopherT AAC Fanatic!

Nov 23, 2012
8,025
6,786

Ok, you were on a perfectly good track for question 20. calculate your total resistance. As you said, your simplified version gives you 4k5 ohms to ground, the parallel block is 1477.8 and the top resistor is 4k. Total is 9977.8 ohms, total current is 50V/9977.8ohms = 0.005011125 A
Then voltage across the bottom resistor (in your simplified version) is 4500 ohms x 0.005011 amps = 22.5501 V.

The voltage across the top resistor (also full current flow of 0.005011527 A) is 4000 ohms x 0.005 amps is 20.0445 V. That leaves 7.4V (of the 50V total) to drop across your parallel section, as you said was a 1500 ohm block and a 100,000 ohm block.

Find your current flow through the 1500 ohm block, you have 7.4054/1500 = 0.004936933 amps (4.93 mA)
Find your current flow though the 100,000 ohm block, you have 7.4054/100000 = 0.000074054 amps (0.074 mA)
Double check that the sum of 4.93633 mA + 0.074 mA = 5.011 mA (close enough with rounding)

Then, the bottom half of the 100k potentiometer (getting 0.000074054 A of current) x 40% of 100k = 2.96216 V (Across the bottom part of the 100k pot.

Add 2.96216V to the Voltage drop across the 4500 ohm resistor (which was calculated at 22.5501V) = 25.51226V

Got it?