Voltage Drop

[MedPR]

I accidentally cut off answer D, but answer D is the correct one.
D: The voltage drop for A is greater than that for B by a factor of 2.




Can someone show me the math please? Here is what I did.

V=IR. Since the resistors are in series, they are additive.

So 24V=I*60Ohms
I=24/60 = 0.4Amps.

Then I did this.

V=0.4A*40Ohms = 16V = Voltage across resistor A. 24-16 = 8

Then V=0.4A*20Ohms = 8 = Voltage across resistor B. 16-8 = 8

Why aren't they equal?

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[aspiringdoc09]

I accidentally cut off answer D, but answer D is the correct one.
D: The voltage drop for A is greater than that for B by a factor of 2.




Can someone show me the math please? Here is what I did.

V=IR. Since the resistors are in series, they are additive.

So 24V=I*60Ohms
I=24/60 = 0.4Amps.

Then I did this.

V=0.4A*40Ohms = 16V = Voltage across resistor A. 24-16 = 8

Then V=0.4A*20Ohms = 8 = Voltage across resistor B. 16-8 = 8

Why aren't they equal?

When you look at resistors in series, the current remains the same through all resistors but the voltage differs (The reverse is true for parallel resistors).

So you have R1 = 20 ohms and R2=40 ohms.

R effective = R1 + R2 = 60 ohms.
Current = V/R effective = 24/60 =0.4. So far you are right.
V across R1 = I*R1 = 0.4*20 = 8 V
V across R2 = I*R2 = 0.4*40 =16 V.

Therefore, R2 voltage drop > R1 voltage drop. You don't need to subtract 8 from 16. You could have subtracted the voltage drop across one resistor from the total battery voltage to get the other resistor's answer (i.e 24 - R1 = 16 V for R2) but you don't need to keep subtracting 8 V from you second resistor.

Just understand that resistors in series have same current and different V and resistors in parallel have different currents and same V

Hope this helps

 

[MedPR]

When you look at resistors in series, the current remains the same through all resistors but the voltage differs (The reverse is true for parallel resistors).

So you have R1 = 20 ohms and R2=40 ohms.

R effective = R1 + R2 = 60 ohms.
Current = V/R effective = 24/60 =0.4. So far you are right.
V across R1 = I*R1 = 0.4*20 = 8 V
V across R2 = I*R2 = 0.4*40 =16 V.

Therefore, R2 voltage drop > R1 voltage drop. You don't need to subtract 8 from 16. You could have subtracted the voltage drop across one resistor from the total battery voltage to get the other resistor's answer (i.e 24 - R1 = 16 V for R2) but you don't need to keep subtracting 8 V from you second resistor.

Just understand that resistors in series have same current and different V and resistors in parallel have different currents and same V

Hope this helps

Is that something we are just supposed to know? Like F=ma?

When I think about that rule, I understand the answer. I don't understand the concept though.

Also, if electricity is like water in that it takes the path of least resistance, why would current ever flow through a stronger resistor when there is a weaker resistor in parallel?

 

[aspiringdoc09]

Is that something we are just supposed to know? Like F=ma?

When I think about that rule, I understand the answer. I don't understand the concept though.

Also, if electricity is like water in that it takes the path of least resistance, why would current ever flow through a stronger resistor when there is a weaker resistor in parallel?

1st Question: Yes. It is imperative that you know that. It becomes really important when you are doing a bunch of problems like the one above but more complicated. Also, there are similar rules when you throw a capacitor in the mix. EK 1001 Physics beats this concept to death. You'll be like, "Hey, resistors in series have same current but I need to figure out the voltage drop across each resistor," and you will easily get your points.

2nd Question: Think about what it means to be parallel. I think they used a good example in my princeton review book. Think about 2 waterfalls (resistors): one wide with little resistance and the other narrow with large resistance. If you drop water (electricity) down to a level with these two waterfalls (resistors), you are bound to have water to flow through both simultaneously than wait to flow through the least resistant waterfall. Because like water, current is a measure of flow rate (Q/t).

This is similar to storm drains on your house. You may have one storm drain that is narrow and one that is wider. However, if it rains, the rain will go down both as oppose to the wider one.

I don't know what review book you are using but TPRH explains this stuff way better than the EK books alone.

 

[MedPR]

At this point I've only looked at NOVA. I plan on reading TBR as well, then EK if I still don't get it. I think I understand what you are saying, thank you.

I do have another question though.

If I have two resistors in series, R1 = 4ohms, R2 = 2 ohms. I tell you that the current through R1 is 2 amps.

So the voltage drop across R1 is 2*4=8V, and the voltage drop across R2 is 2*2=4V. After R2, the wire goes directly to the battery. If the voltage drop across R2 is 4, the voltage between R1 and R2 must be 4V, right? So, if I asked you what the potential from the batter is, it would be 8V+4V = 12V, correct?

 

[SaintJude]

Is that something we are just supposed to know? Like F=ma?

When I think about that rule, I understand the answer. I don't understand the concept though.

Also, if electricity is like water in that it takes the path of least resistance, why would current ever flow through a stronger resistor when there is a weaker resistor in parallel?

Yeah, like F=ma.

Parallel resistors

And voltage will be same across parallel resistors

Series Resistors

Voltage will be, according to changes in resistance, different across resistors in series.

And the "path of least resistance" is a misleading myth. My professor just said to forget it. And like you point out, it will only mislead you. If anything current will proportionately take paths of lesser resistance.

 

[SaintJude]

So, if I asked you what the potential from the batter is, it would be 8V+4V = 12V, correct?

Yeah.

 

[aspiringdoc09]

At this point I've only looked at NOVA. I plan on reading TBR as well, then EK if I still don't get it. I think I understand what you are saying, thank you.

I do have another question though.

If I have two resistors in series, R1 = 4ohms, R2 = 2 ohms. I tell you that the current through R1 is 2 amps.

So the voltage drop across R1 is 2*4=8V, and the voltage drop across R2 is 2*2=4V. After R2, the wire goes directly to the battery. If the voltage drop across R2 is 4, the voltage between R1 and R2 must be 4V, right? So, if I asked you what the potential from the batter is, it would be 8V+4V = 12V, correct?

You got it.