Download Bulbs in series and parallel Bulbs in series Bulbs in series Electron

11/5/2015
Bulbs in series
Bulbs in series and parallel
Given 2 bulbs and a battery, there are 2 different ways that you can
connect the circuit
1.5 V
1.5 V
Compare the brightness of identical bulbs.
How does bulb 1 compare to bulb 2?
a. brighter,
b. dimmer,
120 V
c. same
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Series
Parallel
Bulbs in series
Compare the brightness of identical bulbs.
How does bulb 1 compare to bulb 2?
a. brighter,
b. dimmer,
120 V
c. same
Electron man picture of bulb problem
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Electrons marching around in a line, pushing each other through.
Each electron passes through both bulbs.
Same number of electrons passing by per second at any point on
the circuit. So same current!
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R
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Remember rules:
- No Passing
- No electron man deaths or
births on route
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Series circuit question - current
case 1
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• Think about energy.
• Brightness of bulb → temp → electrical power heating filament (P= IV = I2R)
• All electrons have to go through both bulbs (wired in series), so I is the same in each
• Bulbs identical, so R the same for each
• So power heating filament the same, hence temperature and light out the same.
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Series circuit question - current
case 1
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case 2
All bulbs(R) and batteries (V) are identical.
How does current flow through circuit in case 1 compare
to case 2?
a. Current in case 2 is 1/4 of current in case 1
b. Current in case 2 is 1/2 of current in case 1
c. Current in case 2 is same as current in case 1
d. Current in case 2 is 2 times current in case 1
e. Current in case 2 is 4 times current in case 1
+
All bulbs(R) and batteries (V) are identical.
How does current flow through circuit in case 1 compare
to case 2?
a. Current in case 2 is 1/4 of current in case 1
b. Current in case 2 is 1/2 of current in case 1
c. Current in case 2 is same as current in case 1
d. Current in case 2 is 2 times current in case 1
e. Current in case 2 is 4 times current in case 1
case 2
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11/5/2015
Series circuit question - current
Series circuit question - voltage
case 1
Current for case 1:
How big is voltage drop (difference) across
lefthand bulb (both bulbs identical)?
120 V +
Apply ohm’s law to bulb: I = V/R
a. 60 V,
• So I1 = V/R
b. 120 V,
c. 240 V,
d. 0 V
Current for case 2:
• Apply ohm’s law to whole circuit
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• Rtotal= Rbulb 1 + Rbulb2 = 2 R
• I = V/Rtotal
case 2
?.?? V
• So I2 = V/(2R) = 1/2 I1
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When resistors are strung together in SERIES then:
a) total resistance (Rtotal) is just the sum of the individual
resistances
b) Same current through each resistor
c) Most useful form of power equation usually P = I2R
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Series circuit question - voltage
I
120 V +
How big is voltage drop (difference) across lefthand bulb
(both bulbs have same resistance R)?
a. 60 V,
b. 120 V,
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c. 240 V,
d. 0 V
Series circuit question - power
How will brightness of bulb 1 in circuit A compare to brightness of the same
bulb in the circuit B?
a. brighter A,
b. dimmer in A,
+
120 V
c. same
+
120 V
• Rtotal = 2R
1
• Apply Ohm’s law to whole circuit to find I:
I = Vtotal/(2R)
• Now apply Ohm’s law to left bulb:
Vbulb = IbulbRbulb
=I*R
= Vtotal/(2R) * R
= 120/2
= 60 V
?.?? V
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Circuit B
Circuit A
Series circuit question - power
Bulbs in parallel
How will brightness of bulb 1 in circuit A compare to brightness of the same
bulb in the circuit B?
a. brighter in A,
+
120 V
b. dimmer in A,
c. same
+
120V
1
2
1
2
1
I
Compare the brightness of identical bulbs.
How does bulb 1 compare to bulb 2?
a. brighter,
b. dimmer,
c. same
+
120 V
1
2
I
I = 120/R1
Circuit B
Circuit A
I = 120/(R1 + R2)
Current is smaller in circuit A
 Less electrical power into bulb 1 in circuit A,
P = I2R1
 Less EM power out
 Bulb 1 is less bright in circuit A
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11/5/2015
Thinking like an electron – parallel circuits
Bulbs in parallel
Compare the brightness of identical bulbs.
How does bulb 1 compare to bulb 2?
a. brighter,
b. dimmer,
c. same
No matter what the path, electrons start out with the same
energy from the battery and lose all their energy ( voltage) by
the time they return to outlet.
Vpath1 = Vpath2 = Vbattery
+
120 V
1
If Rwires~ 0  Vpath1 = V1 = Vbattery
Vpath2 = V2 = Vbattery
lots of energy
at start. (Vbattery)
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Path 1
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V2
V1
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Path 2
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No energy left
• Bulbs have same R and same V across them
• Pin = V2/R is the same  brightness the same
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Bulbs in parallel
Bulbs in parallel
B
A
120 V
For bulbs wired in parallel:
• Same voltage across each bulb
• V1 = V2
• (Assumes Rwires ~ 0)
• Most useful form of power equation
is usually P = V2/R
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120 V
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1
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120 V
1
2
V1
2
How does brightness of bulb 1 in circuit A compare to the same bulb in B?
a. brighter,
b. dimmer,
c. same
V2
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Bulbs in parallel
Bulbs in parallel
B
A
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120 V
1
2
Vbattery
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120 V
1
Vbattery
How does size of IB compare to I1 ?
(Bulbs 1 and 2 are identical)
120 V
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IB
1
a)
b)
c)
d)
I1
IB is three times I1
IB is twice I1
IB is half I1
IB is the same as I1
2
How does brightness of bulb 1 in circuit A compare to the same bulb in B?
a. brighter,
b. dimmer,
c. same
P = IV = Vbattery2/R, for both circuits
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Bulbs in parallel
120 V
120 V
How does size of IB compare to I1?
(Bulbs 1 and 2 are identical)
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IB
1
I1
a)
b)
c)
d)
IB is three times I1
IB is twice I1
IB is half I1
IB is the same as I1
Bulb 2
V1
2
• In parallel  V1 = V2 = Vbattery
• V1/R = V2/R  I1 = I2
• IB = I 1 + I 2 = 2 I1
120 V
I2
Bulb 3
Light bulbs wired in series.
If bulb 3 burns out, what happens to the current through bulb 2 ?
a. decreases, b. stays the same, c. increases
V2
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Bulb 2
Answer a. decreases, in fact goes to zero!
Complete circuit is broken  no steady flow of electrons
House wiring
When have desk lamp plugged into wall outlet and plug hair dryer into same outlet,
the two are:
a. in series (each electron flows through lamp then hair dryer, or vice-versa)
b. in parallel (each electron has to choose whether to go through the lamp or
the hair dryer, but never goes through both)
Bulb 3
Light bulbs wired in parallel.
If bulb 3 burns out, what happens to the current through bulb 2?
a. decreases, b. stays the same, c. increases
ans. b. stays the same.
There is still a complete circuit for electrons through bulb 2 and the battery
Voltage drop across bulb is the same,
Resistance of Bulb 2 is the same,
 Current (I2 = V2/R2) is still the same.
Answer: b. in parallel
• Think – if light bulb burns out (filament breaks) does the hairdryer turn off?
• This is one advantage of parallel circuits – Under most conditions different
components operate independently
If bulbs A and B both have a resistance of 20 ohms, what is the total electrical power
converted in the bulbs?
a. 0.056 W,
b. 0.113 W,
c. 0.226 W
1.5 V
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d. 30 W,
e. 60 W
A
We know V and R for each bulb:
B
Pbulb = Vbulb2/R
= 1.52/20
= 0.113 W
Total Power = 2 x 0.113 W
= 0.226 W
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