· Physicsl Adv Physics Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Hr.
U sing a computer simulation to model DC circuits
There are several pHET circuit simulations applets. The one we are going to use now has the following name: Circuit Construction Kit (DC Only), Virtual Lab and has the following website: http://phet.colorado.edu/enlsimulation/circuit-construction-kit-dc-viItual­ lab
You will see the following screen'when you access the website:
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~ Electricity, rvlagnets & Circuits
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Circuit Construction Kit (DC Only), Virtual Lab
Build circuits with r-esistors, light bulbs, batteries, and switches and take meaSut-erTlents v'....
ith laboratorv equipment like tt-Ie realistic ammeter and voltmeter
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Go ahead and click "Run Now!"
When you do this you will now see a screen like this:
This program uses a lot of clicks and drags. You can choose an item from the list and drag in onto the blue area.
By setting up wires, batteries, light bulbs, and resistors, you can create lots of different types of circuits.
Let's get started. pHET DC circuit applet activity, p. 1

· First drag a light bulb onto your workspace. There are connections at the tip and at the threads. Right click on it with your mouse. When you right click on it, there are prompts allowing you to change the resistance and its connections.
Click, "Change Resistance."
Notice you can change the resistance at when you do this. The resistance of the light bulb can vary from
0-100 Ohms. The default is 10 n.
Let's leave it at the default value for now.
Now add two wires and connect them to both connections of the light bulb, one to the tip and the other to the threads.
Notice that wires can be lengthened or shortened with your mouse and dragged around.
The
Why don~t you add a couple of batteries to your workspace?
Notice you can make your batteries vertical or horizontal by pulling on one end.
N ow right click on one of the batteries. This will give you several battery options: "Change Voltage, Change Internal Resistance,
Reverse, and Remove."
Click the, "Change Voltage" prompt. Notice that each battery has a default voltage of 9V with an internal resistance of 0 Ohms. Change BOTH batteries to match the D batteries we use in class. Each of our D batteries has a voltage of 1.5V so change BOTH batteries to have a voltage of 1.5 V.
Now attach the batteries together like the picture on the right: top to bottom, top to bottom. This will give us 3 Volts total, just like if we had 2 D batteries linked together. pHET DC circuit applet activity, p. 2

Now. lefs build a complete circuit. Click on the ammeter icon and then drag one on to your workspace. Also~ drag a switch over to your workspace. Finally, click on the voltmeter and drag one of these to your workspace.
You now have two batteries, a switch, a voltmeter, an ammeter, two wires and a light bulb on your workspace.
Add wires and move things around to build a complete circuit. Attach your voltmeter to your light bulb. Remember that voltmeters are placed across a device. Attach the red probe lead to the tip connection and the black probe lead to the other connector of the light bulb.
Keep the switch open until you have everything set up. Here is what my circuit looks like:
Before we "turn on" the switch, we can determine the current
(what the amn1eter should read) and the reading of the voltmeter.
Using Ohm's Law:
VIR: 3V
=
I (lOO). The light bulb uses all the push from the battery so the voltmeter across the light bulb should read 3 V. The total resistance of the circuit comes from the light bulb and is 10 Ohms. The current then is
3V11 00 = O.3A.
So, your voltmeter should read 3V and your ammeter should read 0.3A. Close the switch and verify that these two meters read the values we think they should read.
I-Iere is a picture of my activated circuit.
Notice the light bulb lights up but it is not very bright. pHET DC circuit applet activity, p. 3

Remember that the brightness of a light bulb is based on the power delivered to the light bulb. Remember that Power delivered to a device VI and is based on the unit, WATT. In this case the power would be
3V*(0.3A) 0.9 Watts.
How could we get a brighter light from the light bulb? Well, we could make the Voltage higher. This would give us a bigger push. Why don't you change the voltage of each battery? Do a right click on each battery and up the voltage of each to some higher value.
Do you see the light bulb light up brighter now?
The other thing we could do is to decrease the resistance of the light bulb. Right click the light bulb and decrease the resistance of the light bulb. As you do this, you should see the light bulb get brighter.
Verify this.
# 1
Let's go back to this basic circuit.
Change each battery's voltage to 9V each.
Make sure the light bulb has 10 Ohms of resistance.
Make sure both batteries are end to end. This will give you 18 V of total voltage.
Keep your voltmeter across the light bulb.
Keep your ammeter in the circuit.
Make sure the switch is closed. circuit #"'1
You should now see a really bright bulb.
Your ammeter should read 1.8 A and your voltmeter should read 18 V. pHET DC circuit applet activity, p. 4

Let ~ s modify the circuit by placing another light bulb in series. We won't change anything else. Break the circuit before the ammeter and add another light bulb in series. You may need to add more wires. It should look like the following schematically.
So, the batteries should each be delivering 9V (total of 18 V) and each bulb should have 10 Ohms of resistance.
Notice some interesting things happen ...
What happens to the light bulbs' brightness levels?
What happens to the ammeter reading?
What happens to the voltmeter reading? circuit #1
You should see the following things on the simulation:
Voltmeter 9V
Ammeter O.9A
Brightness of each bulb gets dimmer
Do you understand why you get these readings?
By adding another light bulb, we have doubled the effective resistance of the circuit.
Consequently, we double the resistance and half the current. Each voltage drop is now 9V as both bulbs are getting half the push. The bulbs get dimmer as each receives half the total power delivered by the batteries. pHET DC circuit applet activity, p. 5

Let ~ s modify the circuit by placing another light bulb in series
(we now have three light bulbs). We won't change anything else. Break the circuit before the ammeter and add another light bulb in series. You may need to add more wires. It should look like the following schematically.
So, the batteries should each be delivering 9V (total of 18 V) and each bulb should have 10 Ohms of resistance.
A
~ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ~l r
Notice some interesting things happen ...
What happens to the light bulbs' brightness levels?
What happens to the ammeter reading?
What happens to the voltmeter reading?
You should see the following things on the simulation:
Voltmeter
=
6V
Ammeter 0.6A
Brightness of each bulb gets dimmer than with two bulbs.
Do you understand why you get these readings?
By adding another light bulb, we have tripled the effective resistance (30
Ohms) of the circuit. With Ohm's
Law V IR, voltage is 18 V from the batteries and the resistance is 30 Ohm. The current from the batteries (reading on ammeter) is 18 V / 30 n
0.6A.
Each voltage drop (reading on the voltmeter) is now 6V as each bulb is getting Y3 of the 18 total volts from the batteries.
The bulbs get dimmer as each receives 1/3 of the total power delivered by the batteries. pHET DC circuit applet activity, p. 6

Let's modify the circuit by placing another light bulb in series (now have four light bulbs). We won't change anything else. Break the circuit before the ammeter and add another light bulb in series. You may need to add more wires. It should look like the following schematically.
So, the batteries should each be delivering 9V and each bulb should have 10 Ohms of resistance.
Notice some interesting things happen ...
What happens to the light bulbs' brightness levels?
What happens to the amnleter reading?
What happens to the voltmeter reading? circuit #1
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You should see the following things on the simulation:
Voltmeter = 4.5V
Ammeter = 0.45A
Brightness of each bulb gets dimmer than with three bulbs.
Do you understand why you get these readings?
By adding another light bulb, we have an effective resistance (40 Ohms) in the circuit.
With Ohm's Law V = IR, voltage is 18 V from the batteries and the resistance is 40 Ohm. The current from the batteries (reading on ammeter) is 18 V /40 n
= 0.45 A.
Each voltage drop (reading on the voltmeter) is now 4.5V as each bulb is getting Y4 of the 18 total volts from the batteries.
The bulbs get dimmer as each receives Y4 of the total power delivered by the batteries.
pHET DC circuit applet activity, p. 7

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\
T
/ total
Next page, let's solve the circuit theoretically. pHET DC circuit applet activity, p. 8

Solving Circuit 2 Theoretically:
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pHET DC circuit appJet activity,
~ p q

Use the computer simulation to build the circuit and then verify your theoretical current, voltage drops, and powers.
Move your voltmeter around to get the voltage drop through each bulb and then the batteries.
Notice the charge is flowing as we get a reading on the ammeter. However, the bulbs don't really appear to light up much. There are a few rays coming from the long bulb but no rays from the round bulb.
Why? There is not enough power to light up the bulbs filaments very much.
Here are the values from my meters:
Current 0.09 A
Voltage drop round bulb Vround
=
0.9 V
Voltage drop long bulb Vlong 3.6 V
Voltage total = 4.5 V
Notice, they match up with our theoretical values from the previous page.
The power to the round bulb is only 0.081 Watts.
This is not enough to energize the filament enough.
The power to the long bulb is 0.324 Watts. It is barely lit. pHET DC circuit applet activity, p. 10

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round
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Next page~ solve the circuit theoretically. pHET DC circuit applet activity, p. 11

Solving Circuit 3 Theoretically:
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Use the computer simulation to build the circuit and verify your theoretical current, voltage drops, and powers.
Move your voltmeter around to get the voltage drop through each bulb and then the batteries.
When you move your voltmeter around, it should read 4.5 V across the long bulb, across the round bulb, and across the batteries.
My simulated circuit tells me the total current
=
0.56 A, the branch current of the round bulb
=
0.45 A, and the branch current of the long bulb = 0.11A.
0.45 A + 0.11 A 0.56 A
Check to see that these currents match up with the theoretical values on the previous page.
My round bulb is fairly bright but not the long bulb.
Does this match up with the powers delivered?
Check the previous page.
The power delivered through the round hulb 0.5063 Watts, not enough to energize the filament to glow.
The power delivered through the long bulb = 2.025 Watts, enough to see the glow.
pHET DC circuit applet activity, p. 13

U sing the computer simulation, take 2 batteries (give them each 3 V for a total of 6V) and add three resistors (RJ 5 11, R2 10 11 and
R3 = 1511) [Make sure to change the resistance of each resistor]
Here is what your meters should read:
I (current) O.2A
Voltage Drop across Rl (VI) = 1 Volt
Voltage Drop across R2 (V2) = 2 Volt
Voltage Drop across R3 (V3) = 3 Volt
Verify that your simulation gives you these values. Move the voltmeter around to verify voltage drops.
pHET DC circuit applet activity, p. 14

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0
...
1=
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7
Oilins
X
n
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.
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( ) 13
~-.-/ .
12
v
Here are the correct ammeter readings: currents (Ib 3.36A, II 2A, h = 0.857 A, and 13 = 0.5A and Ix h 13 1.357A)
Your simulation should verify these ammeter readings.
pHET DC circuit applet activity, p. 15