# Can't figure out question 25 on Voltage Divider worksheet

#### cranberrysky

Joined Feb 15, 2022
54
In Question 25 of the Voltage Divider worksheet, I can't figure out why it can't be an open fault in R3.

One of the resistors in this voltage divider circuit is failed (either open or shorted). Based on the voltage readings shown at each load, determine which one and what type of failure it is:

Resistor R1 has failed shorted.
Follow-up question: note that the voltage at load #2 is not fully 25 volts. What does this indicate about the nature of R1’s failure? Be as specific as you can in your answer.
Notes:
Discuss with your students how they were able to predict R1 was the faulty resistor. Is there any particular clue in the diagram indicating R1 as the obvious problem? Some students may suspect an open failure in resistor R3 could cause the same effects, but there is a definite way to tell that the problem can only come from a short in R1 (hint: analyze resistor R2).
Explain that not all “shorted” failures are “hard” in the sense of being direct metal-to-metal wire connections. Quite often, components will fail shorted in a “softer” sense, meaning they still have some non-trivial amount of electrical resistance.

#### crutschow

Joined Mar 14, 2008
34,037
I can't figure out why it can't be an open fault in R3.
Calculate the node voltages with R3 open.
Do they equal the red values on the diagram?

#### cranberrysky

Joined Feb 15, 2022
54
How can I calculate the hypothetical node voltages without knowing the exact resistor values? I know that the parallel R3 + Load 3 has a total resistance 5/25 of the total circuit resistance, but I don't know how to calculate how exactly it would change if R3 was open, other than knowing it would increase.

#### panic mode

Joined Oct 10, 2011
2,670
try considering part of the circuit too (without V1 and R1). you don't need to know resistor values. you know the ratios and you know their (approximate) value when shorted or open

#### cranberrysky

Joined Feb 15, 2022
54
try considering part of the circuit too (without V1 and R1). you don't need to know resistor values. you know the ratios and you know their (approximate) value when shorted or open
Is it that R2's designed voltage is 9V, and if R3 is open, then R2's voltage should go down, but instead rises to 15.5?

#### panic mode

Joined Oct 10, 2011
2,670
R2/R3 form voltage divider. one side is supposed to be 14V and the other is 5V. if the resistor values are correct changing input voltage will scale output voltage and this is proportional... with this and measured voltages you can deduce if R2 or R3 or both are defective

#### cranberrysky

Joined Feb 15, 2022
54
Is it that the ratio of V3/V2 stays basically the same, when it would increase if R3 was open?

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#### panic mode

Joined Oct 10, 2011
2,670

you can easily derive or lookup voltage divider formula. then replace value of one resistor with a failed value. for short use 0 Ohm, for open use infinity and see what comes out.

#### cranberrysky

Joined Feb 15, 2022
54

you can easily derive or lookup voltage divider formula. then replace value of one resistor with a failed value. for short use 0 Ohm, for open use infinity and see what comes out.

This is as much as I could simplify the circuit. L1 doesn't affect the voltage of the rest of the circuit. I combined R3 and L3 into C. I'm trying to figure out what happens to R2 when C increases some amount, which is what would happen if R3 was open.

If by voltage divider formula, you mean V1=Vt(R1/Rt), it seems like the equation gets pretty complicated because of the L2 branch. First use the voltage divider to get the voltage of everything under R1, then plug that voltage into a new voltage divider formula as Vt to get the voltage of R2. And then see how increasing resistance of C affects V2. I think I got the equation of V2 in terms of C, but it's got three C terms so not very straightforward:

The answer must be simpler than what I've got but I'm stuck.

#### click_here

Joined Sep 22, 2020
548
How can I calculate the hypothetical node voltages without knowing the exact resistor values? I know that the parallel R3 + Load 3 has a total resistance 5/25 of the total circuit resistance, but I don't know how to calculate how exactly it would change if R3 was open, other than knowing it would increase.
Make you calculations simple

First assumption, it is drawing 1A, V=IR, therefore V=R

Now get the resistor values from the desired voltages.

Now play around with the values - If R1 gets really high in value, what happens?
- If R1 gets really Low in value, what happens?
- Same with R2/R3

#### cranberrysky

Joined Feb 15, 2022
54
Make you calculations simple

First assumption, it is drawing 1A, V=IR, therefore V=R

Now get the resistor values from the desired voltages.

Now play around with the values - If R1 gets really high in value, what happens?
- If R1 gets really Low in value, what happens?
- Same with R2/R3
Ok, I plugged in some test values. I made L1, L2, L3, and R3 each draw 1A. With R3 open, the voltage of L2 goes up from design voltage, which is in line with the actual voltage. But the voltage of R2 would down with R3 open (9->7.186), while the actual voltage shows R2's voltage going up (9->15.5). Therefore, the cause of the problem is not R3 being open.

#### click_here

Joined Sep 22, 2020
548
The resistors (R1/2/3) are being used as a voltage reference, so you could assume that the load resistances (L1/2/3) are a lot higher than R1/2/3, say 100k. That would allow the load resistances to change without a significant change of the voltage levels.

By doing this you can justifiably ignore their effects on the circuit.

Now assume R1 = 11 ohms, R2 = 9 ohms, R3 = 5 ohms

You can now look at the affect of an open/short circuit quite easily by making the values either x100 or /100

#### MrChips

Joined Oct 2, 2009
30,466

You can do this without any math.

Measured voltage on load #2 is 24.8V.
Measured voltage on load #3 is 9.3V.
This tells you something about R2 and R3.

Measured voltage on load #1 is 25V.
Measured voltage on load #2 is 24.8V.
This tells you something about R1.

#### cranberrysky

Joined Feb 15, 2022
54
The resistors (R1/2/3) are being used as a voltage reference, so you could assume that the load resistances (L1/2/3) are a lot higher than R1/2/3, say 100k. That would allow the load resistances to change without a significant change of the voltage levels.

By doing this you can justifiably ignore their effects on the circuit.

Now assume R1 = 11 ohms, R2 = 9 ohms, R3 = 5 ohms

You can now look at the affect of an open/short circuit quite easily by making the values either x100 or /100
Oh I see. Well then if any of the resistors increase in resistance, the other two should decrease in voltage. And vice versa.

Out of curiosity, wouldn't this circuit design waste a lot of energy? Since the resistance is so much higher in the loads, most of the current is just going through the resistors instead of the loads.

#### MrChips

Joined Oct 2, 2009
30,466
Out of curiosity, wouldn't this circuit design waste a lot of energy? Since the resistance is so much higher in the loads, most of the current is just going through the resistors instead of the loads.
That is irrelevant. The question is simply testing your analytical skills.

#### click_here

Joined Sep 22, 2020
548
Out of curiosity, wouldn't this circuit design waste a lot of energy? Since the resistance is so much higher in the loads, most of the current is just going through the resistors instead of the loads.
It sure would!

(But remember that we chose 1A just to make the calculation easy)

You'll see this sort of thing (choosing a value so it doesn't affect the rest of the circuit) all the time - Not this extreme though! One example I can think of is the voltage bias section in a common emitter amplifier: You'd usually choose at least a 10:1 ratio

#### cranberrysky

Joined Feb 15, 2022
54
It sure would!

(But remember that we chose 1A just to make the calculation easy)

You'll see this sort of thing (choosing a value so it doesn't affect the rest of the circuit) all the time - Not this extreme though! One example I can think of is the voltage bias section in a common emitter amplifier: You'd usually choose at least a 10:1 ratio
Ok, thanks!