# Simple DC Circuits

## Homework Statement

1)Rank the brightness of the bulbs.
2) What happens to the brightness of the bulbs when A is unscrewed and removed.
3) A is unscrewed, what happens to current through 3,4,5.

Red circles are bulbs and the black circles are reference points.

## The Attempt at a Solution

1) I know that B=C, but I am stuck on trying to figure out if A is brighter , the same, or less than the brightness of the others.

A is in parallel with B and C. Total resistance if each bulb was 1 resistance would equal then 2/3.

How is the current being split up? Is it being split evenly at the intersection of the center left?

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Here are some hints:

1) What is the path of least resistance?

2) Does the potential across both B and C change any?

3) 5 shouldn't get any current, is that supposed to be in there? Did the brightness change across B and C any?

1) A has less resistance so more current would go through it, so would it be A > B=C?

2) Potential wouldn't change, but wouldn't total resistance go up when A is removed. V=IR The current would drop and cause the bulbs to get dimmer? But then all of the current is now going through the two bulbs. :(

3) It says 5 in the question and picture is right, assuming it just wants me to say 0. Same hangup as in 2 to answer this.

cepheid
Staff Emeritus
Gold Member
1) A has less resistance so more current would go through it, so would it be A > B=C?
Yeah.

2) Potential wouldn't change, but wouldn't total resistance go up when A is removed. V=IR The current would drop and cause the bulbs to get dimmer? But then all of the current is now going through the two bulbs. :(
Sure, the TOTAL current drawn from the battery would decrease because the EFFECTIVE resistance the battery "sees" across its terminals would increase. But that's not what determines the brightness of bulbs B and C. What determines the brightness of bulbs B and C is the current actually GOING THROUGH each of them. Has that changed?

I think I understand.

Question 4) was if A is put back in and C is removed what happens to the brightness of each bulb and the current through 3,4,5.

So B and C = no brightness and current in 4 = 0

Current in 3 = 5

Voltage is the same and resistance is the same at bulb A so the current would be the same through 3 and 5 making the Bulb not change.

:)?

Yes, that's right. Bulbs B and C stay the same brightness and have same current through them. The total current, however, changes when A leaves because the battery no longer has to put out so much current to maintain the potential. The total current changes, the actual current in B and C does not.

You can see this very easily with Kirchoff's voltage law (KVL). If it isn't taught to you, you must learn it on your own because it's the most useful circuit tool (other than KCL) that you'll ever get!