# Potential dividers question help finding R2/R1+R2 ⚡️

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#1
Good Afternoon ☀️I have a question about potential dividers which I would really be grateful if anyone would help me with. I have answered all of the questions but I not feel certain with my solutions and would be grateful if someone could please offer any helpful suggestions or advice to improve upon or correct my workings 😁
I have also attached a diagram.

1. Calculate Vout;

a. Vin = 6 V, R1 = 20 Ω, R2 = 40 Ω
Vout=Vin (R2/R1+R2)
Vout=6*(40/20+40)
Vout=4 V

b. Vin =12V,R1 =50kΩ,R2 =10kΩ
Vout=Vin (R2/R1+R2)
Vout=12(10,000/50,000+10,000)
Vout=2 V

2. Calculate R1 for:
a. Vin = 6V, Vout = 5 V, R2 = 100 Ω
R1=(R2*Vin/Vout)-R2
R1=(100*6/5)-100
R1=20 Ω

b. Vin = 12 V, Vout = 5 V, R2 = 10 kΩ
R1=(R2*Vin/Vout)-R2
R1=(10,000*12/5)-10,000
R1=14,000 Ω or 14 kΩ

3. If an electric device rated 1.5 V 10 W is placed at the terminals X and Y, what would need to be the ratio R2 / R1 + R2 for an input voltage of 6 V?

I am really not too sure where to begin here.

Would i substitute Vin = 6V;
Vout=Vin (R2/R1+R2)
Vout=6V (R2/R1+R2)

Then would I have to find Vout;
or would this be equal to 1.5 W?

If it were;
1.5=6*(R2/R1+R2)
So R2/R1+R2=5/6
R2=5 Ω
R1=1Ω

I do not think that this is correct and I am pulling at straws really so I would appreciate any guidance to get me on the right track. 😁
0
9 months ago
#2
(Original post by Alexandramartis)
Good Afternoon ☀️I have a question about potential dividers which I would really be grateful if anyone would help me with. I have answered all of the questions but I not feel certain with my solutions and would be grateful if someone could please offer any helpful suggestions or advice to improve upon or correct my workings 😁
I have also attached a diagram.

1. Calculate Vout;

a. Vin = 6 V, R1 = 20 Ω, R2 = 40 Ω
Vout=Vin (R2/R1+R2)
Vout=6*(40/20+40)
Vout=4 V

b. Vin =12V,R1 =50kΩ,R2 =10kΩ
Vout=Vin (R2/R1+R2)
Vout=12(10,000/50,000+10,000)
Vout=2 V

2. Calculate R1 for:
a. Vin = 6V, Vout = 5 V, R2 = 100 Ω
R1=(R2*Vin/Vout)-R2
R1=(100*6/5)-100
R1=20 Ω

b. Vin = 12 V, Vout = 5 V, R2 = 10 kΩ
R1=(R2*Vin/Vout)-R2
R1=(10,000*12/5)-10,000
R1=14,000 Ω or 14 kΩ

3. If an electric device rated 1.5 V 10 W is placed at the terminals X and Y, what would need to be the ratio R2 / R1 + R2 for an input voltage of 6 V?

I am really not too sure where to begin here.

Would i substitute Vin = 6V;
Vout=Vin (R2/R1+R2)
Vout=6V (R2/R1+R2)

Then would I have to find Vout;
or would this be equal to 1.5 W?

If it were;
1.5=6*(R2/R1+R2)
So R2/R1+R2=5/6
R2=5 Ω
R1=1Ω

I do not think that this is correct and I am pulling at straws really so I would appreciate any guidance to get me on the right track. 😁
It looks like your early parts are correct (can't say I've checked them all in great detail), but I think you are making it more difficult to understand by always using the full force of the potential divider equation rather than applying what it is telling you. If you have two resistors R1 and R2 set up as a potential divider with potential differences V1 and V2 across them, then the ratio of the reistances will be the same as the ratio of the potential differences, i.e. R1:R2 = V1:V2. In addition, you know that V1 + V2 is equal to the emf of the supply.

So, in question 1(a), R1: R2 = 20:40 = 1:2, so the ratio V1:V2 = 1:2. You know that V1 + V2 = 6 V, so split this 6 V into three equal pieces (2 V each), and there will be one piece across R1 (so V1 = 2 V) and two pieces across R2 (so V2 = 4 V).

In situations where the ratios are easy to deal with, this is almost always quicker and simpler than using the whole equation. I would only use the equation when things get too difficult to do in your head, for example, if the resistors were something like 15 Ω and 47 Ω. Even then, you can see what the answer will approximately be, as you nearly have a 1:3 ratio, so it will help you check that your answer makes sense.

You've got the right idea in the last question, but have gone a bit wrong at the end. You are fine as far as

1.5=6*(R2/(R1+R2))

(don't forget to bracket the (R1 + R2)) but you seem to lose the 1. part of the LHS. You should next have

R2/(R1+R2) = 1.5 / 6 (which is equal to 1/4)

rather than 5/6. Notice that again, you can do this without needing all of the algebra. The fraction R2/(R1 + R2) is simply saying "what fraction of the total resistance do I need over the output resistor?". If the input voltage is 6 V and the output voltage is 1.5 V, and it is easy to see that 1.5 is 1/4 of 6, then the output resistance must be 1/4 of the total resistance, and this is what R2/(R1 + R2) is - the fraction of output resistance (R2) out of the total reistance (R1 + R2).

Even if you were correct that R2/(R1 + R2 ) = 5/6, you cannot conclude that R2 = 5 and R1 + R2 = 6. It would work equally well if R2 = 5000 and R1 + R2 = 6000, for example. If you want to think about it mathematically, you have one equation in two unknowns, so all you have is a relationship between R1 and R2, which is not enough to work out either value. Having said that, I expect that there is another part to the question following this, where you are asked to give values for R1 and R2, which you can do using the power rating of the device, which hasn't come into play yet.
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#3
(Original post by Pangol)
It looks like your early parts are correct (can't say I've checked them all in great detail), but I think you are making it more difficult to understand by always using the full force of the potential divider equation rather than applying what it is telling you. If you have two resistors R1 and R2 set up as a potential divider with potential differences V1 and V2 across them, then the ratio of the reistances will be the same as the ratio of the potential differences, i.e. R1:R2 = V1:V2. In addition, you know that V1 + V2 is equal to the emf of the supply.

So, in question 1(a), R1: R2 = 20:40 = 1:2, so the ratio V1:V2 = 1:2. You know that V1 + V2 = 6 V, so split this 6 V into three equal pieces (2 V each), and there will be one piece across R1 (so V1 = 2 V) and two pieces across R2 (so V2 = 4 V).

In situations where the ratios are easy to deal with, this is almost always quicker and simpler than using the whole equation. I would only use the equation when things get too difficult to do in your head, for example, if the resistors were something like 15 Ω and 47 Ω. Even then, you can see what the answer will approximately be, as you nearly have a 1:3 ratio, so it will help you check that your answer makes sense.

You've got the right idea in the last question, but have gone a bit wrong at the end. You are fine as far as

1.5=6*(R2/(R1+R2))

(don't forget to bracket the (R1 + R2)) but you seem to lose the 1. part of the LHS. You should next have

R2/(R1+R2) = 1.5 / 6 (which is equal to 1/4)

rather than 5/6. Notice that again, you can do this without needing all of the algebra. The fraction R2/(R1 + R2) is simply saying "what fraction of the total resistance do I need over the output resistor?". If the input voltage is 6 V and the output voltage is 1.5 V, and it is easy to see that 1.5 is 1/4 of 6, then the output resistance must be 1/4 of the total resistance, and this is what R2/(R1 + R2) is - the fraction of output resistance (R2) out of the total reistance (R1 + R2).

Even if you were correct that R2/(R1 + R2 ) = 5/6, you cannot conclude that R2 = 5 and R1 + R2 = 6. It would work equally well if R2 = 5000 and R1 + R2 = 6000, for example. If you want to think about it mathematically, you have one equation in two unknowns, so all you have is a relationship between R1 and R2, which is not enough to work out either value. Having said that, I expect that there is another part to the question following this, where you are asked to give values for R1 and R2, which you can do using the power rating of the device, which hasn't come into play yet.
Thank you for your reply and for the breadth fo your explanation. I had not really really thought to use the ratio of the resistances, and that this will be the same as the ratio of the potential differences. I will continue to do so in the future as I understand this to be the simpler and more concise method. I also see that I was wrong to conclude (even though R2/(R1 + R2 ) did not equal 5/6) that R2 = 5 and R1 + R2 = 6 as you have shown the equation could still be true for other values.
I have tried to correct the final question as you have shown below, would this now be correct? 😊

3. If an electric device rated 1.5 V 10 W is placed at the terminals X and Y, what would need to be the ratio R2 / R1 + R2 for an input voltage of 6 V?
Vout=Vin (R2/R1+R2)
1.5=6*(R2/(R1+R2))
R2/(R1+R2) = 1.5 / 6
(R2/(R1+R2))=1/4

Does this fully answer the question (it just seems a little short?) Additionally funnily enough there is not another question so I am uncertain why there was mention to the power rating (should I have used it to find the ratio R2=R1+R2?) Thank you again!👍😁
0
9 months ago
#4
(Original post by Alexandramartis)
Thank you for your reply and for the breadth fo your explanation. I had not really really thought to use the ratio of the resistances, and that this will be the same as the ratio of the potential differences. I will continue to do so in the future as I understand this to be the simpler and more concise method. I also see that I was wrong to conclude (even though R2/(R1 + R2 ) did not equal 5/6) that R2 = 5 and R1 + R2 = 6 as you have shown the equation could still be true for other values.
I have tried to correct the final question as you have shown below, would this now be correct? 😊

3. If an electric device rated 1.5 V 10 W is placed at the terminals X and Y, what would need to be the ratio R2 / R1 + R2 for an input voltage of 6 V?
Vout=Vin (R2/R1+R2)
1.5=6*(R2/(R1+R2))
R2/(R1+R2) = 1.5 / 6
(R2/(R1+R2))=1/4

Does this fully answer the question (it just seems a little short?) Additionally funnily enough there is not another question so I am uncertain why there was mention to the power rating (should I have used it to find the ratio R2=R1+R2?) Thank you again!👍😁
Yes, that is right, but as I say, you really don't need to use the full potential divider equation. The ratio of the output voltage to input voltage is 1/4 (easy to see, or simplify from 1.5/6), so this will also be the fraction of the output resistance to the total resistance, and so this fraction, which by definition is R2/(R1 + R2) = 1/4.
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#5
(Original post by Pangol)
Yes, that is right, but as I say, you really don't need to use the full potential divider equation. The ratio of the output voltage to input voltage is 1/4 (easy to see, or simplify from 1.5/6), so this will also be the fraction of the output resistance to the total resistance, and so this fraction, which by definition is R2/(R1 + R2) = 1/4.
Ah, ok I understand. I was just using the full equation again to ensure that I do not miss out on marks hypothetically by showing my working as comprehensively as possible. Thank you again 😊👍
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