# Physics Mrs. Coyle.  Kirchhoff’s Rules  Series Circuits  Equivalent Resistance  Voltage Drop Across Resistors  Brightness of Bulbs in a Series Circuit.

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Physics Mrs. Coyle

 Kirchhoff’s Rules  Series Circuits  Equivalent Resistance  Voltage Drop Across Resistors  Brightness of Bulbs in a Series Circuit

 There is one current path.  All resistors have the same current.

 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

 1 st Rule: (Junction Theorem): At a junction (node), current in= current out  2 nd Rule: (Loop Theorem): In a closed loop the sum of the voltages is zero.

 In a series circuit the total voltage drop across the resistors equals the sum of the individual voltages. V = V 1 + V 2 + V 3

 If the battery’s voltage is 12V and the voltage across R 1 is 5 V, and across R 2 is 4V, find the voltage across R 3.  Answer: 3V

V = V 1 + V 2 + V 3 Using Ohm’s Law: IR eq = IR 1 +IR 2 +IR 3 Equivalent resistance R eq = R 1 + R 2 + R 3

 If the battery’s voltage is 12V and R 1 = 1 Ω R 2 = 2 Ω R 3 = 3 Ω  Find the equivalent resistance.  Find the current.  Find the voltage across each resistor.  Answer: 6 Ω, 2A, 2V, 4V, 6V

 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.

 P= I V  P=I 2 R  P=V 2 / R

 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

 Parallel Circuits  Equivalent Resistance  Brightness of Light Bulb  Combination Circuits

 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

I = I 1 + I 2 + I 3 V =V 1 =V 2 =V 3 Using Ohm’s Law: V/R eq = V/R 1 +V/R 2 + V/R 3 Equivalent Resistance: 1/R eq = 1/R 1 +1/R 2 + 1/R 3

 Find the R eq, I’s.  How does R eq compare with each R? Ans: 0.55 Ω, I= 22A, (12A, 6A, 4A) =1Ω =3Ω =2Ω 12V

 Why should you not plug in too many appliances in the same outlet in a home?

 Ans: 11 Ω, 1.8A, V 1 =9V, V 2 =11V, I 2 =1.1A, I 3 =0.7A =20V =10Ω =15Ω =5Ω

http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/DC/00123.png

 R eq 1 = 71.4 Ω  R eq 2 = 127.3 Ω  R eq = 198.7 Ω  I=0.12A  V 1 = 8.6V  V 2 = 15.3V

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