Solving Series circuits How to Whiteboards A Series

Solving Series circuits • How to • Whiteboards

A Series Circuit has many voltages, but only one current Note that the total voltage drops add to source voltage A 1 1 V 1 = IR = (3 A)(1 ) = 3 V 2 V 2 = IR = (3 A)(2 ) = 6 V V 3 = IR = (3 A)(5 ) = 15 V 24 V A 2 5 A 3 How to solve series: 1. Find the total resistance (1 + 2 + 5 = 8 ) 2. Find the one current (I = V/R = (24 V)/(8 ) = 3 A - all the ammeters read 3 A) 3. Use V = IR to find voltages (voltage drops)

I R 1 A 1 R 2 A 2 R 3 • Analogy - ping pong balls - electrons • All start moving at once/speed of impulse, speed of particles • Resistance anywhere along the tube limits flow everywhere

Whiteboards 1|2|3|4

What is the Total resistance? 5 7 11 20. 0 V R tot = R 1 + R 2 + R 3. . . = 5 + 7 + 11 = 23

What is the reading on both ammeters? (The one current) (answer with 3 SF) 5 7 A 2 11 A 1 20 V I = V/R = (20 V)/(23 ) =. 8696 =. 870 A

What do the voltmeters read? (3 SF) V 1 V 2 5 7 11 20. 0 V V = IR V 1 = (5 )(. 8696 A) = 4. 35 V V 2 = (18 )(. 8696 A) = 15. 7 V 4. 35 V, 15. 7 V

What do the voltmeters read? (3 SF) V 1 V 2 11 5 35 V R = 41 , I =. 8537 A V = IR V 1 = (5 )(. 8537 A) = 4. 27 V V 2 = (20 )(. 8537 A) = 17. 1 V V 3 = (13 )(. 8537 A) = 11. 1 V 4. 27 V, 17. 1 V, 11. 1 V 9 13 V 3 3

What do the voltmeters read? (3 SF) V 1 V 2 500 200 90 110 V 3 120 V R = 1030 , I =. 1165 A V = IR V 1 = (500 )(. 1165 A) = 58. 3 V V 2 = (90 )(. 1165 A) = 10. 5 V V 3 = (330 )(. 1165 A) = 38. 4 V 58. 3 V, 10. 5 V, 38. 4 V 130

What is the value of the mystery resistor if V 1 reads 3. 17 V (3 SF) V 1 5 7 ? ? ? 20 V I = V/R = (3. 17 V)/(5 )=. 634 A Rtot = V/I = (20 V)/(. 634 A) = 31. 55 Rtot = 5 + 7 + ? ? = 31. 55 ? ? = 19. 5

V 3 A 3