Voltage question that is really annoying me!

[flodhi1]

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So my understanding from my Physics 2 lecture and from review notes/books is that In series you have one pathway thus the current is the same when passing through each resistor but voltage drops/ changes. In a parallel you have different pathways and the voltage will be the same for each pathway but the current will change. Is that correct? and if so why is it that the answer to Q1 is B and Q2 is B. It seems with the changing of resistance in one answer is changing current (Q1) while the other is changing Voltage (Q2). Am I conceptually not understand something? thank you for your time!

Q1) A simple circuit contains a battery and two resistors. The resistors are connected in parallel. If one of the resistors were removed, which of the following would be true?
A)The total current in the circuit would increase.
B)The total current in the circuit would decrease.
C)The equivalent resistance of the circuit would decrease.
D)The total voltage drop across the circuit would decrease.

Q2) What can we say about the potential difference across a resistor if the internal resistance of the battery is no longer negligible?
A)The voltage drop across Resistor is higher than in the ideal case.
B)The voltage drop across Resistor is lower than in the ideal case.
C)The voltage drop across Resistor is the same as in the ideal case, but the current through it is higher.
D)The voltage drop across Resistor is the same as in the ideal case, but the current through it is lower.

Okay my attempt to understand it was
Q2) The internal resistor would be added thus it would be in series with the battery. Thus the Req would increase, the current would be the same
V= IR so if I is the same R increases... Voltage should increase? I know that sounds really ******ed and I'm sure that's wrong...
Q1) So with this I assumed that if ONE resistor is removed it goes from a parallel resistance to a circuit resistance. thus to SAME current and voltage difference!/drop so thats why I picked D! but I realize that they are asking about the entire circuit not the resistor so okay maybe I messed up? any elaborations..... Any way all the help would be appreciated. 😀

 

[ilovemcat]

So my understanding from my Physics 2 lecture and from review notes/books is that In series you have one pathway thus the current is the same when passing through each resistor but voltage drops/ changes. In a parallel you have different pathways and the voltage will be the same for each pathway but the current will change. Is that correct? and if so why is it that the answer to Q1 is B and Q2 is B. It seems with the changing of resistance in one answer is changing current (Q1) while the other is changing Voltage (Q2). Am I conceptually not understand something? thank you for your time!

Q1) A simple circuit contains a battery and two resistors. The resistors are connected in parallel. If one of the resistors were removed, which of the following would be true?
A)The total current in the circuit would increase.
B)The total current in the circuit would decrease.
C)The equivalent resistance of the circuit would decrease.
D)The total voltage drop across the circuit would decrease.

Q2) What can we say about the potential difference across a resistor if the internal resistance of the battery is no longer negligible?
A)The voltage drop across Resistor is higher than in the ideal case.
B)The voltage drop across Resistor is lower than in the ideal case.
C)The voltage drop across Resistor is the same as in the ideal case, but the current through it is higher.
D)The voltage drop across Resistor is the same as in the ideal case, but the current through it is lower.

Okay my attempt to understand it was
Q2) The internal resistor would be added thus it would be in series with the battery. Thus the Req would increase, the current would be the same
V= IR so if I is the same R increases... Voltage should increase? I know that sounds really ******ed and I'm sure that's wrong...
Q1) So with this I assumed that if ONE resistor is removed it goes from a parallel resistance to a circuit resistance. thus to SAME current and voltage difference!/drop so thats why I picked D! but I realize that they are asking about the entire circuit not the resistor so okay maybe I messed up? any elaborations..... Any way all the help would be appreciated. 😀

Usually with these types of questions, the way I approach it is by picking a few numbers.

Question 1 says you have two resistors in parallel. What I'd do is make up some numbers for both. Let's say both resistors are 2 ohms. Originally, the equivalent resistance is: 1/R = 1/2 + 1/2 or 1 ohm. But if you take the other resistor and all you're left with is just 1 - then that 1 resistor is in a series. It's equivalent resistance is just the resistance of THAT resistor (since it's the only one). So it's resistance is 2 ohms. (It increases by x2). Assuming Voltage remains constant and resistance increases by a factor of 2 - what happens to current? Well V=IR. Because current (I) and resistance (R) are inversely proportional, an increase in resistance by 2 must mean current decreases by a factor of 2. B is the answer.

Question 2. The way I like to think of internal resistance is, as just that ...another resistor. If internal resistance isn't negligable, it's using up some of the voltage source (just like all resistors do). And because a portion of the voltage source is being used up, there's less voltage to share with the other resistors in the circut. Therefore the other resistors get a smaller portion of the remaining voltage source, i.e. the voltage drop is lower in this case. B is the answer.

 

[flodhi1]

Usually with these types of questions, the way I approach it is by picking a few numbers.

Question 1 says you have two resistors in parallel. What I'd do is make up some numbers for both. Let's say both resistors are 2 ohms. Originally, the equivalent resistance is: 1/R = 1/2 + 1/2 or 1 ohm. But if you take the other resistor and all you're left with is just 1 - then that 1 resistor is in a series. It's equivalence resistance is just the resistance of THAT resistor (since it's the only one). So it's resistance is 2 ohms. (It increases by 1/2). Assuming Voltage remains constant and resistance increases by 1/2 - what happens to current? Well V=IR. Because current (I) and resistance (R) are inversely proportional, an increase in resistance by 1/2 must mean current decreases by a factor of 2. B is the answer.

Question 2. The way I like to think of internal resistance is, as just that ...another resistor. If internal resistance isn't negligable, it's using up some of the voltage source (just like all resistors do). And because a portion of the voltage source is being used up, there's less voltage to share with the other resistors in the circut. Therefore the other resistors get a smaller portion of the remaining voltage source, i.e. the voltage drop is lower in this case. B is the answer.

Thanks that made perfect sense! but one last thing concerning Question 1) How do you know if you can assume or when to assume that the Voltage remains constant? that is the main thing I am struggling with.
I just keep telling myself oh if its parallel VOLTAGE IS CONSTANT but if it's series the current in CONSTANT. Thus in question 1) I assumed since it's going to be series with the removal of the resistor the current would be the same.

 

[ilovemcat]

Thanks that made perfect sense! but one last thing concerning Question 1) How do you know if you can assume or when to assume that the Voltage remains constant? that is the main thing I am struggling with.
I just keep telling myself oh if its parallel VOLTAGE IS CONSTANT but if it's series the current in CONSTANT. Thus in question 1) I assumed since it's going to be series with the removal of the resistor the current would be the same.

I see what you're confusing. Okay, for circuit problems, the voltage source of the battery remains constant. Usually that's the case 100% of the time, unless we're told otherwise. The battery is our voltage source.

Throughout the circuit, we're consuming this voltage source via the resistors in the circuit.

Look, here's an example of a circuit with 4 resistors in a series:

Voltage of Battery = 100V

Resistor 1 = 5 ohm
Resistor 2 = 3 ohm
Resistor 3 = 2 ohm
Resistor 4 = 10 ohm

^--------- These 4 resistors each have a voltage drop of their own, depending on their equivalent resistance. This is different from the voltage source of the battery. What you can say is that the combined voltage drops of each of the resistors WILL add up to the voltage of the battery source.

In order to find these voltage drops, we need to find the equivalent resistance of all these resistors FIRST. Why? Because we don't know the current that's passing through them. Look at your equation for adding resistors in a series. Req = R1 + R2 + R3 + R4. Simply, we just add them up: 5 + 3 + 2 + 10 ohm. Our equivalent resistance is 20 ohms. Now that we found the equivalent resistance, we divide this by the voltage SOURCE to find the current passing through the resistors.

V = IR (V = 100V and R = 20 ohms), therefore
I = 5 Amps

Great, so we found the current that's being supplied to our resistors. Let's look again from the beginning:

Voltage of Battery = 100V
.
.
5 Amps is traveling from voltage source to R1
.
.
Resistor 1 = 5 ohm (V = IR ==> V = 5A x 5ohm)
Voltage Drop = 25V
.
.
5 Amps is traveling from R1 to R2
.
.
Resistor 2 = 3 ohm (V=IR ==> V= 5A x 3ohm)
Voltage Drop = 15V
.
.
5 Amps is traveling from R2 to R3
.
.
Resistor 3 = 2 ohm (V=IR ==> V=5A x 2 ohm)
Voltage Drop = 10V
.
.
5 Amps is traveling from R3 to R4
.
.
Resistor 4 = 10 ohm (V=IR ==> V=5A x 10 ohm)
Voltage Drop = 50V

The sum of all the voltage drops should equal the original voltage of our battery source. If you add them up, you'll see they add to 100V ...equal to the voltage of our battery.


For resistors in parallel - the reason why we take the "voltage drop" as opposed to current is because 2 resistors in parallel could have very different currents passing through them (just like resistors in a series could have very different voltage drops). BEFORE the current passes through the resistors in parallel, it's 5Amps. As it travels between the two, one resistor might have 3Amps traveling through it - the other 2 Amps. As the current exits the resistor in parallel they combine to equal 5Amps once again ....traveling to the next resistor.

Hopefully this makes the whole circuit thing much more understandable. Sorry if I confused you more, lol.

 

[flodhi1]

Hopefully this makes the whole circuit thing much more understandable. Sorry if I confused you more, lol.

Dude wow you're amazing it made perfect sense. I get it now I was so ******ed I looked at it as the Voltage from the source (Emf) instead of looking it as a voltage drop upon each resistor. You're freaking awesome!!!!!!!!!!!!!!! 👍

 

[ilovemcat]

Dude wow you're amazing it made perfect sense. I get it now I was so ******ed I looked at it as the Voltage from the source (Emf) instead of looking it as a voltage drop upon each resistor. You're freaking awesome!!!!!!!!!!!!!!! 👍

No problem haha. I struggled with the same exact thing at one point, so I knew exactly the problem you were having. 👍

 

[ThirdEye]

Nice circuit review!

 

[Rabolisk]

flodhi, I would definitely try to understand exactly what it means to say that current is CONSTANT in series and voltage is CONSTANT in parallel. What it means is that current through each resistor in series is the SAME. It means that the voltage drop across any resistor in parallel is the SAME. Also, for Q1, even if you didn't know that removing parallel resistors increases equivalent resistance of the circuit, you could logically reason through it. First, A and C are equivalent answers, because if equivalent resistance decreases, then total current would increase. You can rule those out. Out of B and D, all you have to realize is that the voltage drop across the entire circuit is equal to the emf, which does not change. B is the correct answer.