TPR - Electric Circuits About BULBS Burning Out

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
jHustle

jHustle

Membership Revoked
Removed
10+ Year Member
Joined
May 14, 2012
Messages
233
Reaction score
3
Members don't see this ad.
Question:

Three identical light bulbs are connected to a battery, as shown:


d1KUw.png


What will happen if the middle bulb burns out?

A. The other two bulbs will go out.
B. The light intensity of the other two bulbs will decrease, but they won't go out.
C. The light intensity of the other two bulbs will increase.
D. The light intensity of the other two bulbs will remain the same.

Can someone please explain to me the answer and their reasoning? Thanks in advance

edit: pic added, mspaint ftw
 
Sort by date Sort by votes
jHustle said:
Question:

Three identical light bulbs are connected to a battery, as shown:


d1KUw.png


What will happen if the middle bulb burns out?

A. The other two bulbs will go out.
B. The light intensity of the other two bulbs will decrease, but they won't go out.
C. The light intensity of the other two bulbs will increase.
D. The light intensity of the other two bulbs will remain the same.

Can someone please explain to me the answer and their reasoning? Thanks in advance

edit: pic added, mspaint ftw
Click to expand...

is it C?
 
Upvote 0 Downvote
I thought C, then changed my mind to D. Reason being, the voltage doesn't change (V1=V2=V3 etc. in parallel). Resistance R of the individual bulb doesn't change either. So from V = IR, current through the bulb doesn't change. The brightness of the bulb depends on the current that flows through it.
 
Last edited:
Upvote 0 Downvote
N1st said:
I thought C, then changed my mind to D. Reason being, the voltage doesn't change (V1=V2=V3 etc. in parallel). Resistance R of the individual bulb doesn't change either. So from V = IR, current through the bulb doesn't change. The brightness of the bulb depends on the current that flows through it.
Click to expand...

so what is the correct answer?

when there's 3 light bulbs, the equivalent resistance of the circuit is lower cuz they are in parallel, but when u take 1 bulb away, the resistance is higher. higher resistance make more heat, and more heat=more light, hence answer C.
 
Upvote 0 Downvote
N1st said:
I thought C, then changed my mind to D. Reason being, the voltage doesn't change (V1=V2=V3 etc. in parallel). Resistance R of the individual bulb doesn't change either. So from V = IR, current through the bulb doesn't change. The brightness of the bulb depends on the current that flows through it.
Click to expand...

oh.. I forgot about that! Good point!!

I was first thinking that total current must stay the same, thus the individual currents going through the remaining resistors would increase.
However, now that you brought it up I realized that actually the overall current decreases since the voltage in the circuit stays the same, and the total resistance decreases.

Thanks for that N1st! 👍
 
Upvote 0 Downvote
N1st said:
I thought C, then changed my mind to D. Reason being, the voltage doesn't change (V1=V2=V3 etc. in parallel). Resistance R of the individual bulb doesn't change either. So from V = IR, current through the bulb doesn't change. The brightness of the bulb depends on the current that flows through it.
Click to expand...

Yep, it's D.

Let's say the circuit is connected to a 9-V battery. Let's also say each light bulb has resistance of 9 ohms. Originally, the equivalent resistance (Req) is then 3 ohms. Using V=IReq, and rearranging, the current through the circuit is 9/3 = 3 A. Given that each bulb is identical, the current will be divided in 3 equal ways, making each bulb get 1 A.

In the second scenario, the Req is 4.5 ohms. Voltage is still the same at 9 V. So, 9/4.5 = 2 A. There are 2 amps running through the circuit. Bulbs 1 and 2 are identical, so the current splits evenly in two ways. Each bulb receives 1 A.

Since brightness is dependent on the current running through the bulb (not on heat produced), and the bulbs in each scenario are receiving the same current (1 A), then the brightness should remain the same.

This is the idea behind wiring systems in your house. Wiring devices in parallel allows more than one appliance to be turned on without affecting the performance of another. It also prevents the entire circuit from going out if one light bulb blows.
 
Last edited:
Upvote 0 Downvote
icecoldstar said:
so what is the correct answer?

when there's 3 light bulbs, the equivalent resistance of the circuit is lower cuz they are in parallel, but when u take 1 bulb away, the resistance is higher. higher resistance make more heat, and more heat=more light, hence answer C.
Click to expand...

as N1st said, the brightness of the bulb depends on the current that goes through it. If I remember correctly intensity is proportional to power, and P=IV.
Hence, since current increases, so does intensity.
 
Upvote 0 Downvote
icecoldstar said:
so what is the correct answer?

when there's 3 light bulbs, the equivalent resistance of the circuit is lower cuz they are in parallel, but when u take 1 bulb away, the resistance is higher. higher resistance make more heat, and more heat=more light, hence answer C.
Click to expand...

The answer is D. If you want to think of it in terms of equivalent resistance, when you remove the bulb, the equivalent resistance of the circuit is increased, but the voltage is unchanged. V = IR, so there is less current. How much less current? The amount that makes it so that each light bulb gets the exact same amount as it did when there were three light bulbs.
 
Upvote 0 Downvote
The answer is C. Yes, the voltage remains the same across, but the current does not.

The answer is fairly intuitive.


Consider this. If I connected a small 9v battery to 100000000000 lamps in parallel, I highly doubt any of them would experience enough resistance to even glow (visibly).

Now, in a blind rage, I smash all but two lamps. Do you see the light?
 
Upvote 0 Downvote
SSerenity said:
The answer is C. Yes, the voltage remains the same across, but the current does not.

The answer is fairly intuitive.


Consider this. If I connected a small 9v battery to 100000000000 lamps in parallel, I highly doubt any of them would experience enough resistance to even glow (visibly).

Now, in a blind rage, I smash all but two lamps. Do you see the light?
Click to expand...

Disclaimer: I'm a former electrical/computer engineer.

I like your thinking, but it is wrong in this case because these MCAT questions assume ideal current and voltage sources, which have zero output impedance. That means a 9V battery on the MCAT could supply infinite current. In reality that's not possible, but hey this is the MCAT's world, we just live in it.

The fallacy in your intuition is that you assume that there is a current limit on a small battery. For this question, there is no limit.

Let's do some math.

Assumptions:

N = number of parallel branches
R1 = resistance of each resistor in parallel
V = voltage of battery
Req = resistance of entire parallel portion of circuit
Ieq = current thru entire parallel portion of circuit

Req = 1 / (N * 1/R1) = R1/N

Ieq = V/Req = V/(R1/N) = VN/R1.

The current thru each individual branch of the parallel circuit by Kirchoff's current law is I1 = Ieq/N = (VN/R1) / (N) = V/R1.

Thus, you can see that the individual current thru each parallel branch does NOT depend on N. It doesnt matter if N = 3 or N = 3 million.

Lets plug in some real numbers to your hypothetical scenario:

N = 100 billion
R1 = 10 ohms
V = 9 V

Scenario #1: All 100 billion lamps are plugged in.
Req = 1 / (N * 1/R1) = R1/N = 10 ohms/100 billion = 10 exp -9
Ieq = 9 V / 10 exp -9 = 9 x 10 exp 9
I = Ieq / N = (9 x 10 exp 9) / (10 exp 10) = 0.9 A

Scenario #2: All 100 billion lamps are taken out of the circuit except for 2
Req = R1/N = 10 ohms/2 = 5 ohms
Ieq = 9V / 5 ohms = 1.8 A
I = Ieq/N = 1.8 A / 2 = 0.9A

The correct answer is D

QED
 
Upvote 0 Downvote
There are a lot of erroneous answers on here. The answer is NOT C. It is D.

The resistance in each bulb REMAINS THE SAME (both before and after the middle bulb is taken out). Why? because resistance is a property of the bulb. For example, if a bulb has a resistance of 2 ohms, then it will continue to have a resistance of 2 ohms regardless of what circuit you put it in. So in this case, taking out the middle bulb will have no effect on the INTRINSIC resistances of the other two bulbs.

Furthermore, the current that flows through each bulb will remain the same. This is because V= IR. We know that the voltage will not change, and we just reasoned that the resistance through each bulb will not change. Thus, I must also be the same through each bulb. If V, I and R are all the same, then the intensity wont change.

What does change? The TOTAL resistance changes, which causes the TOTAL current to change. Everything else stays the same. PM for a more detailed explanation.
 
Upvote 0 Downvote
That's good you set these people straight in their two year old discussion.

It also seems that they settled on the correct answer anyways.

I hope anyone asking questions here has taken the test by now and will likely not PM you 😉
 
Upvote 0 Downvote
You must log in or register to reply here.

Similar threads

M
How representative is the AAMC FL? 8 weeks out from the test and using the AAMC to decide whether to reschedule
Replies
0
Views
1K
mmssjj
M
K
Please Explain to me Like I'm 5: Electric Field/ Magnetic Field
Replies
0
Views
9K
krazzydhoom
K
D
  • Question Question
Concept about circuits from TBR
Replies
2
Views
892
deleted647690
D
K
  • Question Question
Physics Circuits
Replies
5
Views
2K
PlsLetMeIn21
P
A
  • Question Question
When to use which equation of Power (Circuits): P=I^2*R or P=V^2/R ?
Replies
3
Views
6K
autumn123
A
Top