2: Resistor Circuits. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 1 / 13. 2: Resistor Circuits

Transcription

1 and E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 1 / 13

2 Kirchoff s Voltage Law and The five nodes are labelled A, B, C, D, E wheree is the reference node. Each component that links a pair of nodes is called a branch of the network. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 2 / 13

3 Kirchoff s Voltage Law and The five nodes are labelled A, B, C, D, E wheree is the reference node. Each component that links a pair of nodes is called a branch of the network. Kirchoff s Voltage Law (KVL) is a consequence of the fact that the work done in moving a charge from one node to another does not depend on the route you take; in particular the work done in going from one node back to the same node by any route is zero. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 2 / 13

4 Kirchoff s Voltage Law and The five nodes are labelled A, B, C, D, E wheree is the reference node. Each component that links a pair of nodes is called a branch of the network. Kirchoff s Voltage Law (KVL) is a consequence of the fact that the work done in moving a charge from one node to another does not depend on the route you take; in particular the work done in going from one node back to the same node by any route is zero. KVL: the sum of the voltage changes around any closed loop is zero. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 2 / 13

5 Kirchoff s Voltage Law and The five nodes are labelled A, B, C, D, E wheree is the reference node. Each component that links a pair of nodes is called a branch of the network. Kirchoff s Voltage Law (KVL) is a consequence of the fact that the work done in moving a charge from one node to another does not depend on the route you take; in particular the work done in going from one node back to the same node by any route is zero. KVL: the sum of the voltage changes around any closed loop is zero. Example: V DE +V BD +V AB +V EA = 0 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 2 / 13

6 Kirchoff s Voltage Law and The five nodes are labelled A, B, C, D, E wheree is the reference node. Each component that links a pair of nodes is called a branch of the network. Kirchoff s Voltage Law (KVL) is a consequence of the fact that the work done in moving a charge from one node to another does not depend on the route you take; in particular the work done in going from one node back to the same node by any route is zero. KVL: the sum of the voltage changes around any closed loop is zero. Example: V DE +V BD +V AB +V EA = 0 Equivalent formulation: V XY = V XE V YE = V X V Y for any nodes X andy. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 2 / 13

7 Kirchoff s Current Law and Wherever charges are free to move around, they will move to ensure charge neutrality everywhere at all times. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 3 / 13

8 Kirchoff s Current Law and Wherever charges are free to move around, they will move to ensure charge neutrality everywhere at all times. A consequence is Kirchoff s Current Law (KCL) which says that the current going into any closed region of a circuit must equal the current coming out. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 3 / 13

9 Kirchoff s Current Law and Wherever charges are free to move around, they will move to ensure charge neutrality everywhere at all times. A consequence is Kirchoff s Current Law (KCL) which says that the current going into any closed region of a circuit must equal the current coming out. KCL: The currents flowing out of any closed region of a circuit sum to zero. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 3 / 13

10 Kirchoff s Current Law and Wherever charges are free to move around, they will move to ensure charge neutrality everywhere at all times. A consequence is Kirchoff s Current Law (KCL) which says that the current going into any closed region of a circuit must equal the current coming out. KCL: The currents flowing out of any closed region of a circuit sum to zero. Green: I 1 = I 7 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 3 / 13

11 Kirchoff s Current Law and Wherever charges are free to move around, they will move to ensure charge neutrality everywhere at all times. A consequence is Kirchoff s Current Law (KCL) which says that the current going into any closed region of a circuit must equal the current coming out. KCL: The currents flowing out of any closed region of a circuit sum to zero. Green: I 1 = I 7 Blue: I 1 +I 2 +I 5 = 0 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 3 / 13

12 Kirchoff s Current Law and Wherever charges are free to move around, they will move to ensure charge neutrality everywhere at all times. A consequence is Kirchoff s Current Law (KCL) which says that the current going into any closed region of a circuit must equal the current coming out. KCL: The currents flowing out of any closed region of a circuit sum to zero. Green: I 1 = I 7 Blue: I 1 +I 2 +I 5 = 0 Gray: I 2 +I 4 I 6 +I 7 = 0 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 3 / 13

13 KCL Example and The currents and voltages in any linear circuit can be determined by using KCL, KVL and Ohm s law. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 4 / 13

14 KCL Example and The currents and voltages in any linear circuit can be determined by using KCL, KVL and Ohm s law. Sometimes KCL allows you to determine currents very easily without having to solve any simultaneous equations: How do we calculate I? E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 4 / 13

15 KCL Example and The currents and voltages in any linear circuit can be determined by using KCL, KVL and Ohm s law. Sometimes KCL allows you to determine currents very easily without having to solve any simultaneous equations: How do we calculate I? KCL: 1+I +3 = 0 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 4 / 13

16 KCL Example and The currents and voltages in any linear circuit can be determined by using KCL, KVL and Ohm s law. Sometimes KCL allows you to determine currents very easily without having to solve any simultaneous equations: How do we calculate I? KCL: 1+I +3 = 0 = I = 2A E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 4 / 13

17 KCL Example and The currents and voltages in any linear circuit can be determined by using KCL, KVL and Ohm s law. Sometimes KCL allows you to determine currents very easily without having to solve any simultaneous equations: How do we calculate I? KCL: 1+I +3 = 0 = I = 2A Note that herei ends up negative which means we chose the wrong arrow direction to label the circuit. This does not matter. You can choose the directions arbitrarily and let the algebra take care of reality. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 4 / 13

18 and and : Components that are connected in a chain so that the same current flows through each one are said to be in series. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 5 / 13

19 and and : Components that are connected in a chain so that the same current flows through each one are said to be in series. R 1,R 2,R 3 are in series and the same current always flows through each. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 5 / 13

20 and and : Components that are connected in a chain so that the same current flows through each one are said to be in series. R 1,R 2,R 3 are in series and the same current always flows through each. Within the chain, each internal node connects to only two branches. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 5 / 13

21 and and : Components that are connected in a chain so that the same current flows through each one are said to be in series. R 1,R 2,R 3 are in series and the same current always flows through each. Within the chain, each internal node connects to only two branches. R 3 andr 4 are not in series and do not necessarily have the same current. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 5 / 13

22 and and : Components that are connected in a chain so that the same current flows through each one are said to be in series. R 1,R 2,R 3 are in series and the same current always flows through each. Within the chain, each internal node connects to only two branches. R 3 andr 4 are not in series and do not necessarily have the same current. : Components that are connected to the same pair of nodes are said to be in parallel. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 5 / 13

23 and and : Components that are connected in a chain so that the same current flows through each one are said to be in series. R 1,R 2,R 3 are in series and the same current always flows through each. Within the chain, each internal node connects to only two branches. R 3 andr 4 are not in series and do not necessarily have the same current. : Components that are connected to the same pair of nodes are said to be in parallel. R 1,R 2,R 3 are in parallel and the same voltage is across each resistor (even thoughr 3 is not close to the others). E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 5 / 13

24 and and : Components that are connected in a chain so that the same current flows through each one are said to be in series. R 1,R 2,R 3 are in series and the same current always flows through each. Within the chain, each internal node connects to only two branches. R 3 andr 4 are not in series and do not necessarily have the same current. : Components that are connected to the same pair of nodes are said to be in parallel. R 1,R 2,R 3 are in parallel and the same voltage is across each resistor (even thoughr 3 is not close to the others). R 4 andr 5 are also in parallel. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 5 / 13

25 Resistors: Voltage Divider and V X = V 1 +V 2 +V 3 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 6 / 13

26 Resistors: Voltage Divider and V X = V 1 +V 2 +V 3 = IR 1 +IR 2 +IR 3 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 6 / 13

27 Resistors: Voltage Divider and V X = V 1 +V 2 +V 3 = IR 1 +IR 2 +IR 3 = I(R 1 +R 2 +R 3 ) E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 6 / 13

28 Resistors: Voltage Divider and V X = V 1 +V 2 +V 3 V 1 V X = = IR 1 +IR 2 +IR 3 = I(R 1 +R 2 +R 3 ) IR 1 I(R 1 +R 2 +R 3 ) E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 6 / 13

29 Resistors: Voltage Divider and V X = V 1 +V 2 +V 3 V 1 V X = = IR 1 +IR 2 +IR 3 = I(R 1 +R 2 +R 3 ) = IR 1 I(R 1 +R 2 +R 3 ) R 1 R 1 +R 2 +R 3 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 6 / 13

30 Resistors: Voltage Divider and V X = V 1 +V 2 +V 3 V 1 V X = = IR 1 +IR 2 +IR 3 = I(R 1 +R 2 +R 3 ) = IR 1 I(R 1 +R 2 +R 3 ) R 1 R 1 +R 2 +R 3 = R 1 R T wherer T = R 1 +R 2 +R 3 is the total resistance of the chain. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 6 / 13

31 Resistors: Voltage Divider and V X = V 1 +V 2 +V 3 V 1 V X = = IR 1 +IR 2 +IR 3 = I(R 1 +R 2 +R 3 ) = IR 1 I(R 1 +R 2 +R 3 ) R 1 R 1 +R 2 +R 3 = R 1 R T wherer T = R 1 +R 2 +R 3 is the total resistance of the chain. V X is divided intov 1 : V 2 : V 3 in the proportionsr 1 : R 2 : R 3. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 6 / 13

32 Resistors: Voltage Divider and V X = V 1 +V 2 +V 3 V 1 V X = = IR 1 +IR 2 +IR 3 = I(R 1 +R 2 +R 3 ) = IR 1 I(R 1 +R 2 +R 3 ) R 1 R 1 +R 2 +R 3 = R 1 R T wherer T = R 1 +R 2 +R 3 is the total resistance of the chain. V X is divided intov 1 : V 2 : V 3 in the proportionsr 1 : R 2 : R 3. Approximate Voltage Divider: IfI Y = 0, thenv Y = R A R A +R B V X. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 6 / 13

33 Resistors: Voltage Divider and V X = V 1 +V 2 +V 3 V 1 V X = = IR 1 +IR 2 +IR 3 = I(R 1 +R 2 +R 3 ) = IR 1 I(R 1 +R 2 +R 3 ) R 1 R 1 +R 2 +R 3 = R 1 R T wherer T = R 1 +R 2 +R 3 is the total resistance of the chain. V X is divided intov 1 : V 2 : V 3 in the proportionsr 1 : R 2 : R 3. Approximate Voltage Divider: IfI Y = 0, thenv Y = R A R A +R B V X. IfI Y I, thenv Y R A R A +R B V X. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 6 / 13

34 Resistors: Current Divider and resistors all share the samev. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 7 / 13

35 Resistors: Current Divider and resistors all share the samev. I 1 = V R 1 = VG 1 whereg 1 = 1 R 1 is the conductance ofr 1. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 7 / 13

36 Resistors: Current Divider and resistors all share the samev. I 1 = V R 1 = VG 1 whereg 1 = 1 R 1 is the conductance ofr 1. I X = I 1 +I 2 +I 3 = VG 1 +VG 2 +VG 3 = V(G 1 +G 2 +G 3 ) E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 7 / 13

37 Resistors: Current Divider and resistors all share the samev. I 1 = V R 1 = VG 1 whereg 1 = 1 R 1 is the conductance ofr 1. I X = I 1 +I 2 +I 3 = VG 1 +VG 2 +VG 3 = V(G 1 +G 2 +G 3 ) I 1 I X = VG 1 V(G 1 +G 2 +G 3 ) = G 1 G 1 +G 2 +G 3 = G 1 G P whereg P = G 1 +G 2 +G 3 is the total conductance of the resistors. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 7 / 13

38 Resistors: Current Divider and resistors all share the samev. I 1 = V R 1 = VG 1 whereg 1 = 1 R 1 is the conductance ofr 1. I X = I 1 +I 2 +I 3 = VG 1 +VG 2 +VG 3 = V(G 1 +G 2 +G 3 ) I 1 I X = VG 1 V(G 1 +G 2 +G 3 ) = G 1 G 1 +G 2 +G 3 = G 1 G P whereg P = G 1 +G 2 +G 3 is the total conductance of the resistors. I X is divided intoi 1 : I 2 : I 3 in the proportionsg 1 : G 2 : G 3. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 7 / 13

39 Resistors: Current Divider and resistors all share the samev. I 1 = V R 1 = VG 1 whereg 1 = 1 R 1 is the conductance ofr 1. I X = I 1 +I 2 +I 3 = VG 1 +VG 2 +VG 3 = V(G 1 +G 2 +G 3 ) I 1 I X = VG 1 V(G 1 +G 2 +G 3 ) = G 1 G 1 +G 2 +G 3 = G 1 G P whereg P = G 1 +G 2 +G 3 is the total conductance of the resistors. I X is divided intoi 1 : I 2 : I 3 in the proportionsg 1 : G 2 : G 3. Special case for only two resistors: I 1 : I 2 = G 1 : G 2 = R 2 : R 1 I 1 = R 2 R 1 +R 2 I X. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 7 / 13

40 Equivalent Resistance: and We know thatv = V 1 +V 2 +V 3 = I(R 1 +R 2 +R 3 ) = IR T So we can replace the three resistors by a single equivalent resistor of valuer T without affecting the relationship between V and I. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 8 / 13

41 Equivalent Resistance: and We know thatv = V 1 +V 2 +V 3 = I(R 1 +R 2 +R 3 ) = IR T So we can replace the three resistors by a single equivalent resistor of valuer T without affecting the relationship between V and I. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 8 / 13

42 Equivalent Resistance: and We know thatv = V 1 +V 2 +V 3 = I(R 1 +R 2 +R 3 ) = IR T So we can replace the three resistors by a single equivalent resistor of valuer T without affecting the relationship between V and I. Replacing series resistors by their equivalent resistor will not affect any of the voltages or currents in the rest of the circuit. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 8 / 13

43 Equivalent Resistance: and We know thatv = V 1 +V 2 +V 3 = I(R 1 +R 2 +R 3 ) = IR T So we can replace the three resistors by a single equivalent resistor of valuer T without affecting the relationship between V and I. Replacing series resistors by their equivalent resistor will not affect any of the voltages or currents in the rest of the circuit. However the individual voltages V 1, V 2 andv 3 are no longer accessible. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 8 / 13

44 Equivalent Resistance: and Similarly we known thati = I 1 +I 2 +I 3 = V(G 1 +G 2 +G 3 ) = VG P. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 9 / 13

45 Equivalent Resistance: and Similarly we known thati = I 1 +I 2 +I 3 = V(G 1 +G 2 +G 3 ) = VG P. SoV = IR P wherer P = 1 G P = 1 G 1 +G 2 +G 3 = 1 1/R /R /R 3 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 9 / 13

46 Equivalent Resistance: and Similarly we known thati = I 1 +I 2 +I 3 = V(G 1 +G 2 +G 3 ) = VG P. SoV = IR P wherer P = 1 G P = 1 G 1 +G 2 +G 3 = 1 1/R /R /R 3 We can use a single equivalent resistor of resistancer P without affecting the relationship between V andi. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 9 / 13

47 Equivalent Resistance: and Similarly we known thati = I 1 +I 2 +I 3 = V(G 1 +G 2 +G 3 ) = VG P. SoV = IR P wherer P = 1 G P = 1 G 1 +G 2 +G 3 = 1 1/R /R /R 3 We can use a single equivalent resistor of resistancer P without affecting the relationship between V andi. Replacing parallel resistors by their equivalent resistor will not affect any of the voltages or currents in the rest of the circuit. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 9 / 13

48 Equivalent Resistance: and Similarly we known thati = I 1 +I 2 +I 3 = V(G 1 +G 2 +G 3 ) = VG P. SoV = IR P wherer P = 1 G P = 1 G 1 +G 2 +G 3 = 1 1/R /R /R 3 We can use a single equivalent resistor of resistancer P without affecting the relationship between V andi. Replacing parallel resistors by their equivalent resistor will not affect any of the voltages or currents in the rest of the circuit. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 9 / 13

49 Equivalent Resistance: and Similarly we known thati = I 1 +I 2 +I 3 = V(G 1 +G 2 +G 3 ) = VG P. SoV = IR P wherer P = 1 G P = 1 G 1 +G 2 +G 3 = 1 1/R /R /R 3 We can use a single equivalent resistor of resistancer P without affecting the relationship between V andi. Replacing parallel resistors by their equivalent resistor will not affect any of the voltages or currents in the rest of the circuit. R 4 andr 5 are also in parallel. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 9 / 13

50 Equivalent Resistance: and Similarly we known thati = I 1 +I 2 +I 3 = V(G 1 +G 2 +G 3 ) = VG P. SoV = IR P wherer P = 1 G P = 1 G 1 +G 2 +G 3 = 1 1/R /R /R 3 We can use a single equivalent resistor of resistancer P without affecting the relationship between V andi. Replacing parallel resistors by their equivalent resistor will not affect any of the voltages or currents in the rest of the circuit. R 4 andr 5 are also in parallel. Much simpler - although none of the original currentsi 1,, I 5 are now accessible. CurrentI S and the three node voltages are identical. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 9 / 13

51 Equivalent Resistance: and For parallel resistorsg P = G 1 +G 2 +G 3 or equivalently R P = R 1 R 2 R 3 = 1 1/R /R /R 3. These formulae work for any number of resistors. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 10 / 13

52 Equivalent Resistance: and For parallel resistorsg P = G 1 +G 2 +G 3 or equivalently R P = R 1 R 2 R 3 = 1 1/R /R /R 3. These formulae work for any number of resistors. For the special case of two parallel resistors R P = 1 1/R /R 2 = R 1R 2 R 1 +R 2 ( product over sum ) E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 10 / 13

53 Equivalent Resistance: and For parallel resistorsg P = G 1 +G 2 +G 3 or equivalently R P = R 1 R 2 R 3 = 1 1/R /R /R 3. These formulae work for any number of resistors. For the special case of two parallel resistors R P = 1 1/R /R 2 = R 1R 2 R 1 +R 2 ( product over sum ) If one resistor is a multiple of the other Suppose R 2 = kr 1, then R P = R 1R 2 R 1 +R 2 = kr2 1 (k+1)r 1 = k k+1 R 1 = (1 1 k+1 )R 1 E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 10 / 13

54 Equivalent Resistance: and For parallel resistorsg P = G 1 +G 2 +G 3 or equivalently R P = R 1 R 2 R 3 = 1 1/R /R /R 3. These formulae work for any number of resistors. For the special case of two parallel resistors R P = 1 1/R /R 2 = R 1R 2 R 1 +R 2 ( product over sum ) If one resistor is a multiple of the other Suppose R 2 = kr 1, then R P = R 1R 2 R 1 +R 2 = kr2 1 (k+1)r 1 = k k+1 R 1 = (1 1 k+1 )R 1 Example: 1kΩ 99kΩ = kω = ( ) kω E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 10 / 13

55 Equivalent Resistance: and For parallel resistorsg P = G 1 +G 2 +G 3 or equivalently R P = R 1 R 2 R 3 = 1 1/R /R /R 3. These formulae work for any number of resistors. For the special case of two parallel resistors R P = 1 1/R /R 2 = R 1R 2 R 1 +R 2 ( product over sum ) If one resistor is a multiple of the other Suppose R 2 = kr 1, then R P = R 1R 2 R 1 +R 2 = kr2 1 (k+1)r 1 = k k+1 R 1 = (1 1 k+1 )R 1 Example: 1kΩ 99kΩ = kω = ( ) kω Important: The equivalent resistance of parallel resistors is always less than any of them. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 10 / 13

56 Simplifying Resistor and Many resistor circuits can be simplified by alternately combining series and parallel resistors. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 11 / 13

57 Simplifying Resistor and Many resistor circuits can be simplified by alternately combining series and parallel resistors. : 2k+1k = 3k E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 11 / 13

58 Simplifying Resistor and Many resistor circuits can be simplified by alternately combining series and parallel resistors. : 2k+1k = 3k : 3k 7k = 2.1k : 2k 3k = 1.2k E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 11 / 13

59 Simplifying Resistor and Many resistor circuits can be simplified by alternately combining series and parallel resistors. : 2k+1k = 3k : 3k 7k = 2.1k : 2k 3k = 1.2k : 2.1k+1.2k = 3.3k E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 11 / 13

60 Simplifying Resistor and Many resistor circuits can be simplified by alternately combining series and parallel resistors. : 2k+1k = 3k : 3k 7k = 2.1k : 2k 3k = 1.2k : 2.1k+1.2k = 3.3k Sadly this method does not always work: there are no series or parallel resistors here. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 11 / 13

61 Non-ideal Voltage Source and An ideal battery has a characteristic that is vertical: battery voltage does not vary with current. Normally a battery is supplying energy so V andi have opposite signs, soi 0. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 12 / 13

62 Non-ideal Voltage Source and An ideal battery has a characteristic that is vertical: battery voltage does not vary with current. Normally a battery is supplying energy so V andi have opposite signs, soi 0. An real battery has a characteristic that has a slight positive slope: battery voltage decreases as the (negative) current increases. Model this by including a small resistor in series. V = V B +IR B. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 12 / 13

63 Non-ideal Voltage Source and An ideal battery has a characteristic that is vertical: battery voltage does not vary with current. Normally a battery is supplying energy so V andi have opposite signs, soi 0. An real battery has a characteristic that has a slight positive slope: battery voltage decreases as the (negative) current increases. Model this by including a small resistor in series. V = V B +IR B. The equivalent resistance for a battery increases at low temperatures. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 12 / 13

64 Summary and Kichoff s Voltage and Current Laws and components Voltage and Current Dividers Battery Internal Resistance For further details see Hayt et al. Chapter 3. E1.1 Analysis of Circuits ( ) Resistor Circuits: 2 13 / 13