# Light bulb brightness (series and parallel)

1. Feb 26, 2006

### jimmy_neutron

Hi, my question relates to light bulb brightness in a circuit with light bulbs connected in series and in parallel. Ill give you the question and then my argument.

Q1:In this diagram light bulbs A,B, and C are all equal. Rank in order the most bright to the least bright.
A1: I would argue that B and C are equal in brightness and are both greater in brightness than A because the amount of current running through B and C is equal and greater than A, and since the amount of current running through a bulb determines brightness, A must be the least bright.

Q2: Suppose a wire is connected between points 1 and 2 running through bulb C. What happens to the brightness in all the bulbs
A2: Bulb C would be short circuited and thus go out, but I am not sure what happens to A and B. Please help

Q3: What happens to the current drawn from the battery when the wire is connected
A3: The current drawn from the battery is decreased because there is no current flowing through C?? Please help

Sorry for the long post but please look over my arguments and tell me if im right or wrong. THANKS!!

2. Feb 26, 2006

### Staff: Mentor

If A, B and C are equal, and the total current through A is divided between B and C, then . . . ?

Remove C from the circuit, and what do you have?

The resistance of two parallel resistances is half of one, i.e. R and R are in parallel then the resulting resistance is R/2.

Resistances add in series, i.e. R1 + R2, but in parallel, 1/R = 1/R1 + 1/R2.

3. Feb 26, 2006

### jimmy_neutron

Ok, so for Q1, A > B=C because A has twice the current of B and C
Q2= C does not glow, A=B because the current running through A is equal to the current running through B. Thus A stayed the same brightness, and B's brightness increased.
Q3= if Q2 is correct, then the light bulbs are in series and the ammount of current running through the circuit hasnt changed since C was removed thus the amount of current drawn from the battery is the same?

4. Feb 26, 2006

### Staff: Mentor

Correct.

Let me back up. If a wire bypasses C, i.e. C is shorted, the current will go through the wire, thus also by passing B, which still has resistance. Therefore only A has current through it.

If C was burned out (open circuit), the current would go through A and B, which would have resistance 2R (R = resistance of A and B each). For interest, what is the resistance of B and C in parallel, if resistance = R for each case.

Any time total resistance drops, the current increases. If the total resistance increases, current drops. I = V/R.

5. Feb 26, 2006

### jimmy_neutron

first off thank you very much for your help.
If a wire is connected between 1 and 2, then C is short circuited and there is no light, however current still runs through the wire. Thus A stays the same brightness and both B and C lack light.
the resistance of B and C in parallel is 1/Req = 1/Rb + 1/Rc and B still has resistance as does A. When the wire is connected, C lacks resistance? so then total reistance dropped and the current drawn from the battery increases?

6. Feb 27, 2006

### Staff: Mentor

When C is short-circuited, so is B, because they are in parallel. Then only A provides resistance (R) and draws current.

If two resistances R are connected in parallel, the combined resistance = R/2. So with A (also resistance R) connected in series to the parallel set of B and C, the total resistance would be R + R/2 (initial circuit).

Now short-circuit R/2 and one has R, so current increases.

If C were to break (open circuit), then the circuit would include A in series with B, with the same current since A and B are in series, but the resistance is now R + R = 2R, and the current would decrease.

Its important to remember the significance of open and short circuit, and series vs parallel connections.

Last edited: Feb 27, 2006
7. Feb 27, 2006

### Tom Mattson

Staff Emeritus