Circuits with Light Bulbs in Series ```Circuits with Light Bulbs in Series
Showing Flow Rate (Current) using symbols.
“Starbursts” are shown around light bulbs to indicate relative brightness. Flow rate arrowstails are drawn
above wires to show amount of Flow rate (current).
Let’s just spend a few minutes practice setting up a basic circuit and using the meters.
Part 1: One Light Bulb in Series (Use round light bulb)
This is a basic series circuit. Sometimes it is helpful to move your fingers around the circuit. When you
have one complete pathway only, you have a series circuit. Build circuit 1 and write down the ammeter and
voltmeter readings. Then build the other two circuits and write down those ammeter and voltmeter readings.
The circuit remains the same except for the placement of the voltmeter.
Series circuits activity, physics / adv physics, p. 1
Circuit 1 Ammeter ___________
Circuit 1 Voltmeter ___________
Circuit 2 Ammeter ___________
Circuit 2 Voltmeter ___________
Circuit 3 Ammeter ___________
Circuit 3 Voltmeter ___________
Schematically, all three circuits are the same. Your current may be
slightly different (due to experimental factors) but all three ammeter
Should there be any ‘real’ difference if the ammeter is placed
before the light bulb or afterwards as long as it is placed in
series?
Why shouldn’t there be a difference?
Charge flow is maintained equally at all points in a series circuit!
When we place the voltmeter across the light bulb, we get the voltage drop across the light bulb. Since the
resistances of all of the other devices (wire, ammeter, switch) are VERY small, the only significant
resistance is from the light bulb; the voltage provided by the batteries should equal the voltage drop
across the bulb. Voltage is CONSERVED in a series circuit!
Look at the degree of brightness of the bulb and keep this in mind as we build more circuits.
The light bulb was
very bright
bright
not super bright
barely lit
Use ‘starbursts’ and arrowtails on your circuits.
Series circuits activity, physics / adv physics, p. 2
Part Two: Two Light bulbs in Series (Use Round Light Bulbs)
Now build the circuit 4 and write down the ammeter and voltmeter readings.
Circuit 4 ammeter _______________
Circuit 5 ammeter _______________
Circuit 4 voltmeter _______________
Circuit 5 voltmeter ______________
(I get a current of around 0.2. Yours may be different depending on strength of your batteries and the resistance of the light bulb.
I get a voltage of around 2.4V)
How does this value compare to the current with just one bulb in series (see page 1)?
So what happens to the current as we add another ‘identical’ light bulb in series? (remember, by
Why? Think about Ohm’s Law, V = IR or I = V/R
The effective resistance increases as we add bulbs in series. With added resistance, the current drops given
the same voltage applied (Ohm’s Law, V = I*R , if R goes up, I must go down for V to stay the same).
the same.
Why should Vbatteries = Vacross both lamps (ideally) ? See page 1 and discuss the voltage drop vs the voltage
from the batteries.
Series circuits activity, physics / adv physics, p. 3
Now let’s put the voltmeter across each of the bulbs separately. Keep the circuit the same, just move the
voltmeter.
Now put the voltmeter across just the top
bulb. (I get around 1.2 V.)
Vcircuit 6 or left light bulb = __________ V
Now put the voltmeter across just the
bottom bulb. (I get around 1.2 V here, too.)
Vcircuit 7 or right light bulb = __________ V
Why should Vcircuit 6 or left light bulb = Vcircuit 7 or right light bulb ?
And if we add both of these up, we get ______________. But this should be close to the voltage
supplied by the batteries. Why?
Each voltage drop added up should be approximately equal to the voltage of the batteries or the voltage across both lamps together.
Again, voltage drops should equal (at least ideally) to the voltage supplied by the batteries. And both light bulbs should have
about the same voltage drop as both light bulbs are identical and have approximately the same resistance.
(Key point…) How did the brightness of the two light bulbs in series contrast with just one light bulb?
The light bulbs were very bright
bright
not super bright
barely lit
Remember that brightness of a light bulb is roughly related to power (P= V across the device I through the device).
Calculate the approximate power of the only light bulb in PART ONE: Power = ___________ W
Calculate the approximate power of one of the light bulbs in THIS PART: Power = _________ W
Should the power of each light bulb decrease or increase as you add more bulbs in series? Why?
Series circuits activity, physics / adv physics, p. 4
Part 3: Three identical light bulbs in series (Use round bulbs)
What do you think happens to the current (ammeter reading) when we now have three bulbs in series?
If all three bulbs are identical, should each have a voltage drop that is numerically similar? Why?
Here is the circuit to build. Initially keep the voltmeter around all three round bulbs. Then move the
voltmeter around to measure the batteries and each light bulb separately. Write down your readings.
Ammeter (Current) = ___________
Voltmeter reading (across all three bulbs) = __________
Voltmeter reading (across batteries) = _________
Voltmeter reading (across left light bulb) = __________
Voltmeter reading (across middle light bulb) = __________
Voltmeter reading (across right light bulb) = ___________
b) Vacross all three light bulbs = Vbatteries = Vlamp1 + Vlamp2 + Vlamp3
Check yours:
____________ V = __________ V = ____________V + _____________V + _______________V ????
When I did this part I get:
2.7V ≈ 2.75 V ≈ 0.80V + 0.80V + 1.0V
2.75 V ≈ 2.6 V (close enough!)
I get a current of about 0.18 A
Are the light bulbs pretty dim now?
I see a brightness decrease with three bulbs in series. Let’s look at the electrical power. We can make this
statement:
Powerbattery ≈ Powerlamp1 + Powerlamp2 + Powerlamp3 but since power = VI, we can rewrite this statement:
Vbattery*(I) ≈ Vlamp1*(I) + Vlamp2*(I) + Vlamp3*(I)
(see next page for my example calculations)
Series circuits activity, physics / adv physics, p. 5
For my circuit:
2.75 V (0.18A) ≈ 0.80V(0.18A) + 0.80V(0.18A) + 1.0V(0.18A)
0.495 W ≈ 0.144 W + 0.144 W + 0.180 W
0.495 W ≈ 0.468 W (pretty close!)
Check your powers in a similar way:
With three light bulbs, each light bulb should have a smaller power than what you found on the previous
page (at the top) for the power of each light bulb in the first two parts. Each bulb is a bit dimmer!
Part 4: Two different Light Bulbs in Series
a) In this part you will use two different light bulbs, one of the #14 round bulbs and then one of the ‘long’
(#48) light bulbs. Here are the specs on these two light bulbs. Resistances are approximate:
Bulb Type
Name
#14
#48
Round
Long
Approximate Resistance (Hot)
Approximate Resistance (Cold)
10 Ω
40 Ω
1.2 Ω
4.5 Ω
Voltmeter reading = ___________ (I get around 2.92 for this)
Round bulb brightness: _____________
Long bulb brightness: _____________
(Note: this is very different than the brightness levels of the
bulbs in the circuit on page 2. In that circuit, both bulbs were
identical and were approx the same brightness.)
When I create the circuit, it doesn’t appear that there is any brightness from the round bulb. This would
imply that the circuit is broken. Why is this reasoning incorrect?
Series circuits activity, physics / adv physics, p. 6
Unscrew the round bulb and take it out of the circuit. What happens to the long bulb? Why? Does this
Now switch the round bulb and the long bulb around. Does this make any difference to our observations?
Does it really make any difference where bulbs or resistors are placed in a completely series circuit?
Why/why not?
Now move your voltmeter to construct these circuits:
Vacross both lamps =
_____________
(from previous
page)
Vround bulb =
____________
V long bulb =
____________
Vbatteries =
___________
Verify this statement:
Vacross both lamps ≈ Vround bulb + V long bulb ≈ Vbatteries
Here’s what I get:
Vbatteries = 2.99 V
Vacross both lamps = 2.92 V
Vround bulb = 0.077 V (Interesting!)
Vacross both lamps ≈ Vround bulb + V long bulb ≈ Vbatteries
Mine came out pretty close: 2.92 V ≈ 2.907 V ≈ 2.99 V
Check your voltage statement (equation) here:
Series circuits activity, physics / adv physics, p. 7
V long bulb = 2.83 V
Now let’s look at power (P = VI)
Powersupplied by batteries ≈ Pacross both lamps ≈ Pround bulb + P long bulb
Vsupplied by batteries* (I) ≈ Vacross both lamps* (I) ≈ Vround bulb* (I) + V long bulb* (I)
2.99 V*(0.07 A) ≈ 2.92 V*(0.07 A) ≈ 0.077 V*(0.07 A) + 2.83 V*(0.07 A)
0.209 W ≈ 0.204 W ≈ 0.00539 W + 0.19810 W
0.209 W ≈ 0.204 W ≈ 0.2035 W pretty close
Now calculate your powers…here is some room to work.
Can you make another statement now about why we don’t see the round bulb light up? We know that
there is current flow (why? The current probe tells us and the important fact that the long bulb is lighting up
tells us that we have a complete pathway)
Basically, the round bulb doesn’t have enough electric pressure (i.e. voltage) to energize the filament to a hot
enough temperature to glow. You can relate this to the extremely low power delivered to the round bulb.
Another way to look at this is at the resistances (see the resistance table on the previous page). The voltage
drop is really Ohm’s Law (V = IR). Because the long bulb has approximately four times the resistance as the
round bulb, the voltage drop across the long bulb is approximately four times greater.
And this scenario exists regardless of if we switch the positions of the two light bulbs! Pretty interesting
stuff!
Part 5: Two Long Light Bulbs in Series
Let’s build one more circuit but before we do this, let’s think about it first.
This is the circuit we will build:
But before we build it, I would like your group to calculate the
following given that the long bulbs are supposedly identical
(approximately 40 Ω of “hot” resistance each):
Theoretical voltage drop (left bulb) = ___________ V
Theoretical voltage drop (right bulb) = ____________ V
Theoretical current = ___________ A
Don’t peek at my answers until you give it a try!
Series circuits activity, physics / adv physics, p. 8
Theoretical voltage drop (top bulb) = Theoretical voltage drop (bottom bulb) = 3 V from the batteries / 2 bulbs = 1.5 V each (assuming both
batteries are new and fully charged. Not probably exactly the case!)
Theoretical current = 0.0375 A. Use Ohm’s Law: V=I(Rtotal) >>> 3V = (I)*80 Ω (total resistance) Solve for I to get 0.0375 A
Build the circuit and look at your experiment current and voltage drops (you’ll need to move around your
voltmeter):
Real circuit voltage drop (left bulb) = ___________ V
Real circuit voltage drop (right bulb) = ____________ V
Real circuit current = ___________ A
Here’s what I get for my real circuit:
Vbatteries = 3.008 V (pretty close to what I thought)
Current = 0.048 A (a bit low from my theoretical current but not really too far off)
Voltage drop top bulb = 1.491 V
Voltage drop bottom bulb = 1.436 V
Both voltage drops are pretty close to 1.5 V
Overall conclusions of a series circuit:
Remember, in series, resistances add up numerically and we get a total or equivalent resistance equal to the
sum of all the resistances (Req or R total = R1 + R2 +…Rn) Even though the wires have a small resistance, we
normally don’t add these to the mix because the resistances are low.
As we add more light bulbs, resistance increases. Because R grows, the current drops given a set voltage.
This is different than resistors in parallel but we will look at parallel circuits soon.
Because the brightness is related to power delivered to each bulb, we see the bulbs dim as we add more of
the same bulb in series.
Later, we will look at resistors in series instead of light bulbs. However, because light bulbs are effective
resistors, the same effects occur.
Nice job on this activity. We looked at different combinations of series circuits:
• One round light bulb
• Two round light bulbs
• Three round light bulbs
• One round and one long bulb
• Two long bulbs
Series circuits activity, physics / adv physics, p. 9
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