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Old 08-20-2019, 11:15 PM   #1
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Default Resistors in parallel.

I have a question....
What happens to a circuit with 3 resistors connected in parallel when one of the resistors is open circuited?

A. Circuit resistance increases
B. Circuit current increase
C. Voltage across each of the two remaining resistors increases.
D. Amount of power consumed in the circuit remains the same.

My answer was A. The more resistors you have in parallel, the lower the circuit resistance, so losing one of the parallel resistor would increase the circuit resistance... is this correct?

I am being told the answer is C.
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Old 08-20-2019, 11:48 PM   #2
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Quote:
Originally Posted by MikeJH View Post
I have a question....
What happens to a circuit with 3 resistors connected in parallel when one of the resistors is open circuited?

A. Circuit resistance increases
B. Circuit current increase
C. Voltage across each of the two remaining resistors increases.
D. Amount of power consumed in the circuit remains the same.

My answer was A. The more resistors you have in parallel, the lower the circuit resistance, so losing one of the parallel resistor would increase the circuit resistance... is this correct?

I am being told the answer is C.
You were correct. Voltage in a parallel circuit is equal across each leg.

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Old 08-21-2019, 12:04 AM   #3
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Old 08-21-2019, 01:34 AM   #4
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Quote:
Originally Posted by MikeJH View Post
I have a question....
What happens to a circuit with 3 resistors connected in parallel when one of the resistors is open circuited?

A. Circuit resistance increases
B. Circuit current increase
C. Voltage across each of the two remaining resistors increases.
D. Amount of power consumed in the circuit remains the same.

My answer was A. The more resistors you have in parallel, the lower the circuit resistance, so losing one of the parallel resistor would increase the circuit resistance... is this correct?

I am being told the answer is C.
To discount C, if you have 3 incandescent lamps wired up in your kitchen and one burns out, do the other 2 get brighter?

An incandescent lamp is essentially a resistor. The voltage will stay the same, as they are wired in parallel.

A: Correct by the formula: 1/RT= 1/R1+1/R2+1/R3, etc.. Where RT is the resistance total, and R1, R2, R3, etc are the resistors.
B: I=V/R. If R increases, I will drop
C: See above
D: P=VI. If the voltage stays the same, and the current drops, the power consumed will also drop.
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Old 08-21-2019, 06:46 AM   #5
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Quote:
Originally Posted by glen1971 View Post
To discount C, if you have 3 incandescent lamps wired up in your kitchen and one burns out, do the other 2 get brighter?

An incandescent lamp is essentially a resistor. The voltage will stay the same, as they are wired in parallel.

A: Correct by the formula: 1/RT= 1/R1+1/R2+1/R3, etc.. Where RT is the resistance total, and R1, R2, R3, etc are the resistors.
B: I=V/R. If R increases, I will drop
C: See above
D: P=VI. If the voltage stays the same, and the current drops, the power consumed will also drop.
+1

wow glen You actually explained it better than my instructors in collage did
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Old 08-21-2019, 09:15 AM   #6
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wow glen You actually explained it better than my instructors in collage did
Thanks!
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Old 08-21-2019, 09:19 AM   #7
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A is correct. Now if you had two resistors in parallel that were in series with a third, then the voltage would increase across the third resistor if one of the parallel resistors became open.
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Old 08-21-2019, 06:41 PM   #8
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Now if you had two resistors in parallel that were in series with a third, If the third resistor became open ?
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Old 08-21-2019, 07:11 PM   #9
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Then no current is flowing...the voltage drop would be at the 'open' point.
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Old 08-21-2019, 08:01 PM   #10
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Quote:
Originally Posted by MikeJH View Post
I have a question....
What happens to a circuit with 3 resistors connected in parallel when one of the resistors is open circuited?

A. Circuit resistance increases
B. Circuit current increase
C. Voltage across each of the two remaining resistors increases.
D. Amount of power consumed in the circuit remains the same.

My answer was A. The more resistors you have in parallel, the lower the circuit resistance, so losing one of the parallel resistor would increase the circuit resistance... is this correct?

I am being told the answer is C.
A is correct and C is correct only if there's current flow.

But they failed to include the most important choice.....

E. It would blow up.

I'm genuinely surprised at how many test questions omit this all-important answer........lol.
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Old 08-21-2019, 10:04 PM   #11
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Thanks for the replies everyone. Micromind could you please explain why C would be correct if there is current flow? And A would be correct in both cases (current flow and no current flow?)
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Old 08-21-2019, 10:25 PM   #12
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C could be correct in that the combined voltage drop across 3 resistors would be the same as the combined voltage drop across just two. But, individually, the voltage drop would have to increase given a constant source voltage and only 2 resistors. This would have to be just part of a more complex circuit. Just a simple source and parallel resistors would not support this answer. I think.

But, I would say there insufficient information as presented for a definite answer.

A is certainly true.
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Old 08-21-2019, 11:52 PM   #13
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Ok now I am confused. My (poor) understanding is this...
In a parallel circuit the voltage drop across each component (resistor) is going to be the source voltage. No matter if there are 2 parallel branches or 10, no matter the resistance of each component, no matter the total circuit resistance, difference of potential across each resistor will be equal to the source voltage. The thing that changes is the current draw depending on the resistance of each component. So if losing a resistor (causing resistance to go up) were to cause the voltage across each resistor to go up, that would mean the new voltage across each resistor would now be higher than source voltage which is not possible. I am not arguing against anybody, everyone here is probably smarter than me when it comes to this. Again this is just my understanding and I am just trying to comprehend how C would be possible.
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Old 08-22-2019, 12:13 AM   #14
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No, there is a voltage drop across the resistor.

To calculate the voltage drop across a resistor, remember: Ohm's Law (V = I x R ) is your friend. Find the current flowing through a resistor, then multiply the current in amps by resistance in ohms to find the voltage drop in volts.

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Old 08-22-2019, 12:16 AM   #15
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Micromind has it right but you can’t determine C without more info.

If we have a theoretical voltage source across the resistors, the voltage across them does not change no matter how many resistors there are. But since I=V/R each one is adding V/R more current and the total resistance is going up if we take one away. So A is correct and in this case. V does not change because it’s an ideal voltage source so C is false.

Now say we have a real voltage source with some resistance. Let’s say it’s 1 ohm and each resistor is 6 ohms and we have a 10 V source. So with all 3 we have 2 ohms in series with 1 ohm so current is 3.33 A. Voltage across resistors is V=IR = 3.33 x 2 = 6.67 V across them.

Take one away and now we have 3 ohms in series with 1 ohm or 4 ohms which gives us 2.5 A. 2.5 x 3 ohms = 7.5 V so the voltage drop across them has increased. Then C would be correct. A is still correct too. The values for all the resistors don’t matter. The voltage will always go up if it is a nonideal source.


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Old 08-22-2019, 01:12 AM   #16
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What is meant by theoretical/ideal vs real/nonideal voltage source? Could you give me an example?
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Old 08-22-2019, 01:20 AM   #17
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edit nevermind.

Last edited by MikeJH; 08-22-2019 at 01:23 AM.
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Old 08-22-2019, 01:44 AM   #18
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I’m still having a hard time picturing what you are trying to describe. In my mind, the voltage source is constant. 10 volts. The resistors are constant, 6 ohms. In this equation it seems that current is the variable. If you increase resistance, less current is going to flow since you have the same source voltage and a now higher resistance. If current were the constant then I could understand how it alters the voltage. But isn’t resistance the opposition to current flow meaning more resistance will alter the current but not the voltage?
Maybe I should get some sleep and try this tomorrow with a fresh mind. I’m not sure where my break up in understanding is coming from.
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Old 08-22-2019, 02:41 AM   #19
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hi All
i think we are over thinking again
it is easy if You have only resistors in parallel the total voltage of the supply (battery volts total) will be the volt drop over the parallel circuit (don't matter what the R total is) this will be achieved with the 3rd variable in the equation namely I
the Amps will vary to allow the equation to work out
this will change if you have a series and parallel circuit

that is my humble opinion (how we learned it 700 Years back)
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Old 08-22-2019, 03:46 AM   #20
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if total circuit resistance increases (A) ?
Then the voltage across the resistors can only drop
as less current across a fix ohm must mean less voltage.
So ( C ) simply cannot be so !
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