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PhysicsMrs. Coyle
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Kirchhoff’s Rules Series Circuits
Equivalent Resistance Voltage Drop Across Resistors Brightness of Bulbs in a Series Circuit
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There is one current path.
All resistors have the same current.
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Positive charges are “pumped” by the battery from low to high potential. V>0
When traversing a resistor with the current, there is a decrease in potential. V<0
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1st Rule: (Junction Theorem): At a junction (node), current in= current out
2nd Rule: (Loop Theorem): In a closed loop the sum of the voltages is zero.
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In a series circuit the total voltage drop across the resistors equals the sum of the individual voltages.
V = V1 + V2 + V3
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If the battery’s voltage is 12V and the voltage across R1 is 5 V, and across R2 is 4V, find the voltage across R3 .
Answer: 3V
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V = V1 + V2 + V3
Using Ohm’s Law:IReq = IR1+IR2 +IR3
Equivalent resistanceReq = R1 + R2 + R3
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If the battery’s voltage is 12V and R1 = 1Ω R2 = 2Ω
R3 = 3Ω Find the equivalent
resistance. Find the current. Find the voltage
across each resistor.
Answer: 6Ω, 2A, 2V, 4V, 6V
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The greater the power actually used by a light bulb, the greater the brightness.
Note: the power rating of a light bulb is indicated for a given voltage, at room temperature and the bulb may be in a circuit that does not have that voltage.
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P= I V
P=I2 R
P=V2 / R
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Find the total resistance.
Find the current. Find the power
dissipated in each lamp. Which light bulb will be
the brightest and why? Find the total power. How does the total
power compare to the powers of the individual bulbs.
Ans: 450Ω, 0.027A, 0.18W, 0.036W, 0.109W, 250 Ω, 0.324W
250Ω
150Ω
50Ω
12 V
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Parallel Circuits Equivalent Resistance Brightness of Light Bulb
Combination Circuits
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There is more than one current path.
The voltage across the resistors is the same.
http://www1.curriculum.edu.au/sciencepd/energy/images/energy_ill112.gif
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I = I1 + I2 + I3
V =V1=V2=V3
Using Ohm’s Law:V/Req= V/R1 +V/R2 + V/R3
Equivalent Resistance:1/Req= 1/R1 +1/R2 + 1/R3
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Find the Req , I’s.
How does Req compare with each R?
Ans: 0.55Ω, I= 22A, (12A, 6A, 4A)
=1Ω=3Ω
=2Ω12V
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Why should you not plug in too many appliances in the same outlet in a home?
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Ans: 11 Ω, 1.8A, V1=9V, V2=11V, I2=1.1A, I3=0.7A
=20V
=10Ω
=15Ω
=5Ω
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http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/DC/00123.png
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Req 1 = 71.4Ω
Req 2 = 127.3Ω
Req = 198.7Ω I=0.12A V1 = 8.6V
V2 = 15.3V