Agenda for Today. Elements of Physics II. Resistance Resistors Series Parallel Ohm s law Electric Circuits. Current Kirchoff s laws


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1 Resistance Resistors Series Parallel Ohm s law Electric Circuits Physics 132: Lecture e 17 Elements of Physics II Current Kirchoff s laws Agenda for Today Physics 201: Lecture 1, Pg 1
2 Clicker Question 1: The lightbulbs in the circuits below are identical with the same resistance R.. Which circuit produces more light? (brightness power) a) circuit I b) circuit II c) both the same d) it depends on R Physics 201: Lecture 1, Pg 2
3 Clicker Question 1: Physics 201: Lecture 1, Pg 3
4 Clicker Question 2: What happens to the voltage across the resistor R 4 when the switch is closed? a) increases b) decreases c) stays the same R 1 S V R 3 R 4 R 2 Physics 201: Lecture 1, Pg 4
5 Clicker Question 3: Which resistor has the greatest current going through it? Assume that all the resistors are equal. a) R 1 V b) both R 1 and R 2 equally c) R 3 and R 4 d) R 5 e) all the same Physics 201: Lecture 1, Pg 5
6 Clicker Question 3: V Physics 201: Lecture 1, Pg 6
7 Problem In the circuit shown below, rank in order, from most to least bright, the brightness of bulbs A E. Physics 201: Lecture 1, Pg 7
8 Clicker Question 4: Bulbs A and B in the figure are identical, and both are glowing. Bulb B is removed from its socket. (a) Does the potential difference Delta V 12 between points 1 and 2 increase, stay the same, decrease, or become zero? (a) (b) (c) (d) increase stay the same decrease become zero Physics 201: Lecture 1, Pg 8
9 Kirchhoff s Junction Law For a junction, the law of conservation of current requires that: where the symbol means summation. This basic conservation statement is called Kirchhoff s junction law. Physics 201: Lecture 1, Pg 9
10 Kirchhoff s Loop Law For any path that starts and ends at the same point: The sum of all the potential differences encountered while moving around a loop or closed path is zero. This statement is known as Kirchhoff s loop law. Physics 201: Lecture 1, Pg 10
11 Steps for Using Kirchhoffs Rules Label all currents Label +/ for all elements Emf (regular +/) Resistors current goes from + to  Choose a loop and direction (your choice) Wit Write down voltage drops across resistors Loop rule (use first sign you get to) List voltages Write down current equation Junction rule Physics 201: Lecture 1, Pg 11
12 Loop Rule Physics 201: Lecture 1, Pg 12
13 Example Solve for the currents in all parts of the circuit. Anytime there is a junction current must split!!! +  I 1 is not necessarily through resistor R R 1 = 10 1!!! + A  Current can only I change at a junction!!! 2 I R 2 = Current in a segment w/o a junction must I I I + remain constant! I 1 2 = 5 V = 10 V B Physics 201: Lecture 1, Pg 13
14 Three loops Example I 1 = I 2 + I 3 Bottom loop clockwise from A I 2 (R 1 ) = 0 (1) + I R 2 = 10 3 R 1 = 10 Top loop clockwise from A + A  I 3 (R 2 )  2 I 2 (R 1 )= 0 (2) I 22 = 5 V +  B Outside loop clockwise from A I 1 I (R )  = 0 (3) + I 3 (R 2 )  1 = 0 (3)  1 = 10 V Physics 201: Lecture 1, Pg 14
15 I 2 (R 1 ) = 0 (1) I 3 (R 2 )  2 I 2 (R 1 )= 0 (2) Example I 3 = 1 A I 2 =.5 A 2 I 1 = 1 A +.5 A = 1.5 A I 3 (R 2 )  1 = 0 (3) I R 2 = I = I + I (4)  A R 1 = I 2 2 = 5 V +  B I = 10 V Physics 201: Lecture 1, Pg 15
16 Clicker Question 5: A circuit is built out of resistors and two batteries as shown in the below. According to Kirchhoff's Current Rule: (a) I 1 +I 2 = I 3 (b) I 2 +I 3 = I 1 (c) I 1 I 3 = I 2 (d) I 1 +I 2 = I 3 (e) I 1 +I 3 = I 2 Physics 201: Lecture 1, Pg 16
17 Clicker Question 6: A circuit is built out of resistors and two batteries as shown in the below. According to Kirchhoff's Loop Rule the left loop tells us: (a) 12 V  (3 )I 1 +(6 )1 2 = 0 (b)  12 V  (3 )I 1 +(6 )1 2 = 0 (c) 12 V  (3 )I 1  (6 )1 3 = 0 (d) 12 V  (3 )I 1  (6 )1 1 = 0 (e) 12 V  (3 )I 1  (6 )1 2 = Physics 201: Lecture 1, Pg 17
18 Clicker Question 7: A circuit is built out of resistors and two batteries as shown in the below. According to Kirchhoff's Loop Rule the right loop tells us: (a) 12 V  (3 )I 1 +(6 )1 2 = 0 (b)  12 V  (3 )I 1 +(6 )1 2 = 0 (c) 12 V  (4 )I 2  (6 )1 3 = 0 (d) 12 V  (4 )I 2 + (6 )1 3 = 0 (e)  12 V + (4 )I 2 + (6 )1 3 = Physics 201: Lecture 1, Pg 18
19 Clicker Question 8: A circuit is built out of resistors and two batteries as shown in the below. According to Kirchhoff's Loop Rule the outside loop tells us: (a) 24 V  (3 )I 1  (4 )1 3 = 0 (b) 24 V  (3 )I 1 + (4 )1 3 = 0 (c)  (3 )I 1 + (4 )1 2 = 0 (d)  (3 )I 1  (4 )1 3 = 0 (e) 24 V  (3 )I 1 + (4 )1 3 = Physics 201: Lecture 1, Pg 19
20 Clicker Question 9: A circuit is built out of resistors and two batteries as shown in the below. Find the current I 3. (a) 4.01 A (b) 2.87 A (c) 1.76A (d) 156A 1.56 (e) 0.93 A Physics 201: Lecture 1, Pg 20
21 Clicker Question 9: A circuit is built out of resistors and two batteries as shown in the below V + (4 )I 2 + (6 )1 3 = 0  (3 )I 1 + (4 )1 2 = 0 12 V  (3 )I 1  (6 )1 3 = 0 I 1 +I 2 = I 3 Physics 201: Lecture 1, Pg 21
22 Clicker Question 10: The ammeter reads 3 A. What is I 1? (a) (b) (c) (d) (e) 3 A 2 A 1 A 2.5 A 1.5 A Physics 201: Lecture 1, Pg 22
23 Clicker Question 10: The ammeter reads 3 A. What is I 1? Physics 201: Lecture 1, Pg 23
24 Example: In the following figure, what is the value of the potential at points a and b? Physics 201: Lecture 1, Pg 24
25 Clicker Question 11: Determine the current going through the 12 resistor: (a) 0.7 A (b) 0.9 A (c) 1.2 A A R 1 =12 1 = 6 V (d) 1.5 A (e) 2.0 A 2 = 24 V R 2 =8 Physics 201: Lecture 1, Pg 25
26 Clicker Question 12: The lightbulbs in the circuit are identical. When the switch is closed, what happens? a) both bulbs go out b) intensity of both bulbs increases c) intensity it of both bulbs decreases d) A gets brighter and B gets dimmer e) nothing changes Physics 201: Lecture 1, Pg 26
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