Lecture # 2 Basic Circuit Laws

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1 CPEN 206 Linear Circuits Lecture # 2 Basic Circuit Laws Dr. Godfrey A. Mills gmills@ug.edu.gh Phone: February 5, 206 Course TA David S. Tamakloe CPEN 206 Lecture 2 205_206

2 What is Electrical Network o Electrical network is an interconnection of electric circuit elements (active and passive) or devices. o A circuit (from last Lecture) is defined as a network that provides one or more closed paths. o Elements of electrical circuits are mostly found interconnected in several ways or network topologies. o In a network topology, we study the properties that relate to the placement of elements in the network and the geometric configuration of the network such as nodes, branches, loops and meshes. CPEN 206 Lecture 2 205_206 2

3 What is a Node o A node is a point in a circuit where two or more circuit elements are connected. o In circuit representation, a node is indicated by a dot. o If a connecting/conducting wire connects two nodes, the two nodes constitutes a single node. o We shall see the usefulness of circuit nodes under node analysis techniques in network analysis lecture. V s CPEN 206 Lecture 2 205_206 3

4 What is a Branch o Branch is a path in a circuit that connects two nodes. o It represents any two terminal element such as a voltage source or a resistor. CPEN 206 Lecture 2 205_206 4

5 What is a Loop and a Mesh o A loop is any closed path through a circuit in which a node is not encountered more than once. o A mesh on the other hand is a loop which does not contain any other loops within it. o A loop is considered independent if it contains at least one branch that is not a part of any independent loop. o In a loop arrangement, a voltage that is encountered from to is considered a positive voltage and a voltage encountered from to is negative o Arrows represent voltage differences; they point from low to high voltage o A loop does not need to have circuit elements. o We shall talk more on loops and mesh in lecture 4 under network analysis techniques. CPEN 206 Lecture 2 205_206 5

6 Ohms Law o Georg Ohm (German Physicist) showed (826) by experiment that a linear relationship exists between the current and voltage in an electric circuit. Thus: n voltage across terminals of conductor is directly proportional to the current through it. i.e., o V I, is constant of proportionality o This expression remains true as long as physical conditions of the conductor remains constant. o NB current direction, voltage polarity is important. If current flows from higher to lower potential, voltage becomes positive and if it is reverse, is negative. o Value of may ranges from 0 to infinity. If 0, we have a short circuit, which gives a zero voltage. CPEN 206 Lecture 2 205_206 6

7 Sample Calculations o Question : o (a) A 30V battery is connected to a 5kΩ resistor. Find the current through the resistor, conductance, and the power dissipated. o (b) If the 30V battery source is replaced with a 2mA current source of, find the voltage, conductance, and power. o (c) If the voltage source in (a) is replaced with a sinusoidal source of 20sinπt V, find the current through the resistor and the power dissipated. o (d) An electric iron draws a current of 3A at 240V. Find the resistance of the iron and energy (KWh) it consumes every year if it is used 5hrs in a week. CPEN 206 Lecture 2 205_206 7

8 Sample Calculations o Question 2: o (a) Determine the conductance of a short circuit on a 240V that results in a short circuit current of 500A. o (b) A resistor is rated for 0kΩ, W. What is its maximum voltage and current ratings. o (c) An electric motor in a blender operating from a 240V voltage source takes current of 8A. If the motor has an efficiency of 80%, find the output of power of the motor required to blend. o (d) A 20Ω resistance is connected to an unknown resistance and the two are connected across a 220V DC supply. The power loss in the resistor is 50W. Determine the value of. CPEN 206 Lecture 2 205_206 8

9 Kirchhoff Current Law (KCL) o Gustav obert Kirchhoff (German Physicist) in 847 postulated two basic laws for analyzing circuits. o The laws focus on sum of voltages around a loop and currents entering and leaving a node. o Kirchhoff Current Law (denoted KCL) states that n The algebraic sum of all the currents entering or terminating at any anode in a circuit equals zero at any instant of time. o KCL is a consequence of the conservation of charge. Thus, any charge that enters a node must leave the node since it cannot be stored there. o Since algebraic sum at a node must be zero, its time derivative must also be zero at any instant of time. CPEN 206 Lecture 2 205_206 9

10 Kirchhoff Current Law (KCL) Consider a node connecting several branches: i i 2 i 3 i 2 (t) i 3 (t) i (t) i 4 (t) i 5 (t) i 4 n j i ( t) j 0 o Use reference directions to determine whether currents are entering or leaving the node with no concern about actual current directions CPEN 206 Lecture 2 205_206 0

11 Implications of the KCL o KCL tells us that all of the elements in a single branch carry the same current. o We say these elements are connected in series. Current entering node Current leaving node i i 2 i 2 CPEN 206 Lecture 2 205_206

KCL Example Figure shows a typical automobile electrical distribution system to various circuit elements, which are; headlight, taillight, starter motor, fan, power locks, dashboard panel.

12 KCL Example Figure shows a typical automobile electrical distribution system to various circuit elements, which are; headlight, taillight, starter motor, fan, power locks, dashboard panel. V bat I bat I head I tail I dash The battery must supply adequate current to independently satisfy the requirements of each load. Note that in this case the rated power requirement for each load may be the same or different. CPEN 206 Lecture 2 205_206 2

13 KCL Example Assume the circuit below is a Christmas tree light with 50 light bulbs and each bulb is rated W. Find the source current 20V I s I b CPEN 206 Lecture 2 205_206 3

14 KCL Solution o Find currents through each bulb: I B W/20V 8.3mA o Apply KCL to the top node: I S 50I B 0 o Solve for Input Current I S : I S 50 I B 47mA CPEN 206 Lecture 2 205_206 4

15 Practice Questions o Question : o If there are three inductors L, L2, and L3, arranged in parallel with a voltage V connected across the elements, show that the equivalent inductance is given by the summation of the inverse all the inductances. o Question 2: o If there are three capacitors C, C2, and C3, arranged in parallel with a voltage V connected across the elements, show that the equivalent capacitance is given by the summation of all the capacitances. CPEN 206 Lecture 2 205_206 5

16 Kirchhoff Voltage Law (KVL) o Kirchhoff Voltage Law (denoted KVL) states that n The algebraic sum of all the branch voltages around a any closed loop in a circuit equals zero at all instants of time. o Alternatively, the KVL law can be stated in terms of voltage drops and rises as follows: the sum of voltage rises and drops in a closed loop at any instant of time are equal. o The Kirchhoff voltage law is a consequence of the conservation of energy as voltage in our earlier definition is energy or work per unit charge. o Note: with KVL, polarity of voltages is important. CPEN 206 Lecture 2 205_206 6

17 Kirchhoff Voltage Law (KVL) V V 2 V 3 V V 2 V 3 0 è V V 2 V 3 o What would happen if you forgot (deliberately ignored) the convention that to is positive and went around a loop using to is positive CPEN 206 Lecture 2 205_206 7

18 Implications of the KVL o KVL tells us that any set of elements which are connected at both ends carry the same voltage. o We say these elements are connected in parallel. v a _ v b _ Applying KVL in the clockwise direction, starting at the top: v b v a 0 è v b v a CPEN 206 Lecture 2 205_206 8

19 KVL Example V B V B 20V 50 Bulbs Total V B CPEN 206 Lecture 2 205_206 9

20 KVL Solution o Note that the same current I will flow through the source and each resistor element. o To solve for I, we apply KVL around the loop: n 228I 228I 228I 20V 0 n I 20V/(50 228Ω) 0.5mA o We solve for the voltage across each resistor as: n V I 0.5mA 228W 2.4V o Since the circuit has one source and equal resistor values, the current will be given by: n [source voltage] divided by the [sum of the resistors]. CPEN 206 Lecture 2 205_206 20

21 KVL Application o Applying the KVL to the circuit below gives the following equations: n Loop : ΣV k 0 >> V B I I n Loop 2: ΣV k 0 >> V B2 I 2 2 I 2 3 I CPEN 206 Lecture 2 205_206 2

22 KVL Example 2 o Find the current I flowing through the circuit and the voltage V ab across the circuit. CPEN 206 Lecture 2 205_206 22

23 KVL Solution 2 o Find the current I flowing through the circuit and the voltage V ab across the circuit. o Since we have a single loop, we can apply the KVL to the circuit. o From the loop, we have (note direction of current flow and polarity of voltage): n 5*I 5 5*I n >> 0*I 5 >> I 0.5A n From the circuit diagram, the terminal voltage V ab across the 5ohm and the 0V is given by: o V ab 5*I 0 >> 2.5V CPEN 206 Lecture 2 205_206 23

24 KVL Example 3 o The circuit below has two sources of supply: an independent voltage source and dependent current source. o Find the current Io delivered by the network and the power that is dissipated in 400Ω load. CPEN 206 Lecture 2 205_206 24

25 KVL Solution 3 o From the circuit, we can identify two loops. Lets assign currents in the two loops as Io and I. o We now apply the KVL to loop, we obtain: n 400Io 750Io 750I 80 () n >> 50Io 750I 80 (2) o We next apply the KVL to loop 2, we obtain: n I 0.6Io (3) o Substituting (3) into (2), and solving, we have: n Io 80/600 50mA o The power dissipated in the 400 ohms is given by: n P2 V*Io I 2 (50mA) 2 *400 W o Question: n Find the power in the 750 ohms resistors CPEN 206 Lecture 2 205_206 25

26 Practice Questions 2 o Question : o If there are three inductors L, L2, and L3, arranged in series with voltages of V, V2, and V3, across the elements respectively, show that the equivalent inductance is given by the summation of the all the inductances. o Question 2: o If there are three capacitors C, C2, and C3, arranged in series with voltages of V, V2, and V3, across the elements respectively, show that the equivalent capacitance is given by the summation of the inverse of all the capacitances. CPEN 206 Lecture 2 205_206 26

27 Current Divider Circuit o Given the circuit below, how do we find I and I 2? I I I 2 2 V o From the KCL theorem, we have o I I 2 I o Since voltage across the elements is V, we have: o I V/ and I 2 V/ 2 CPEN 206 Lecture 2 205_206 27

28 CPEN 206 Lecture 2 205_ Current Divider Circuit 2 2 V V V I I I V 2 2 I V I o This is the current divider formula. o It tells us how to divide the source current among the elements in the circuit in direct proportion. o The larger the element the smaller its share.

29 Current Divider Example I I 2.7A 44Ω 360Ω V Find I and I 2 CPEN 206 Lecture 2 205_206 29

30 Current Divider Solution I I 360Ω.7A 44Ω 360Ω 44Ω.7A 44Ω 360Ω A 0.334A CPEN 206 Lecture 2 205_206 30

31 MultiParallel Circuit o Given the circuit below, how do we I, I 2, and I 3? I I 2 I 3 I 2 3 V o From the KCL theorem, we have o I I 2 I 3 I o Since voltage across the elements is V, we have: o I V/ ; I 2 V/ 2 ; I 3 V/ 3 CPEN 206 Lecture 2 205_206 3

32 CPEN 206 Lecture 2 205_ MultiParallel Circuit Analysis V V V V I 3 2 I V

33 General Current Divider Formula Consider a current divider circuit with >2 resistors in parallel: I I I 2 I V V I 2 3 I 3 V 3 I / / / 3 2 / 3 CPEN 206 Lecture 2 205_206 33

34 Current Divider Example 2 I I 2 I 3.67A 44Ω 360Ω 240Ω V Find I CPEN 206 Lecture 2 205_206 34

35 Current Divider Solution 2 eq 72Ω 44Ω 360Ω 240Ω V Ieq.67A 72Ω 20V I 20V 44Ω V 0.833A CPEN 206 Lecture 2 205_206 35

36 Comments on Circuit Divider o It is difficult to simply or directly employ the current divider technique on circuit problem that has multiple circuits arranged in parallel o To solve such multiple circuit problem using the current divider technique will involve the following steps: n reduce or lumped the circuit to a two parallel circuit problem n solve for the parameters of interest n After obtaining values for two circuit systems, go back to lumped circuit and separate it further into parallel circuits again and continue with analysis till you get to the final solution. CPEN 206 Lecture 2 205_206 36

37 Practice Question 3 o Question : n Two coils are connected in parallel and a voltage of 200V is applied to the terminals. Total current supplied is 25A and the power dissipated in one coil is.5kw. Find the resistance of each coil. o Question 2: n Two devices with resistances 20Ω and 30Ω are connected in parallel and the resultant circuit is connected in series to a 0Ω device. If the circuit is driven by a 240V source that delivers 4.54 A, draw the circuit diagram and find the currents through the 20Ω and 30Ω devices. CPEN 206 Lecture 2 205_206 37

38 Voltage Divider Circuit o Consider the two resistors arranged in series with a voltage v(t) across them as illustrated below. How do we find the voltage across each element in the circuit. v(t) 2 v (t) v 2 (t) v ( t) v( t) v 2 ( t) v( t) CPEN 206 Lecture 2 205_206 38

39 Voltage Divider Circuit V Ohms >> V I I V I 2 2 V V V I I I( 2) 2 2 V I 2 2 V o This is the voltage divider formula. o It tells us how to divide the source voltage among the elements in the circuit in direct proportion. o The larger the element, the larger its voltage drop. CPEN 206 Lecture 2 205_206 39

40 Voltage Divider Example Audio Amplifier Speaker O v O (t) v L (t) L CPEN 206 Lecture 2 205_206 40

41 Voltage Divider Solution o Find v L (t) for the following values: v O (t) 5V O 6W L 4W v L v L ( t) v ( t) O O L 4Ω ( t) 5V 4Ω 6Ω L 2V CPEN 206 Lecture 2 205_206 4

42 Voltage Divider Example 2 kω kω kω 0V 5V 2kΩ V 2 2kΩ V 3 kω o Find an equivalent resistance for the network with V 2 across it, then find V 2. CPEN 206 Lecture 2 205_206 42

43 Voltage Divider Solution 2 kω kω 0V 5V 2kΩ V 2 kω V kω 5V kω kω 2 2.5V CPEN 206 Lecture 2 205_206 43

44 Voltage Divider Solution 2 kω kω kω 0V 5V 2.5V 2kΩ 2kΩ V 3 kω V kω 2.5V kω kω 3.25V CPEN 206 Lecture 2 205_206 44

45 Practice Question 4 o Question : o Three resistances 750Ω, 600Ω, and 200Ω are connected in parallel. If the total current is A, find the voltage applied to the terminals and the current in each branch. o Question 2: o Two coils in a circuit are serially connected and a 220V dc voltage is applied to the terminals of the circuit. The resistance of one coil is 20 ohms and the other is unknown. If the power loss in the unknown coil is 50W, find the value of the resistance of the unknown coil. CPEN 206 Lecture 2 205_206 45

Lecture #3. Review: Power

Lecture #3. Review: Power Lecture #3 OUTLINE Power calculations Circuit elements Voltage and current sources Electrical resistance (Ohm s law) Kirchhoff s laws Reading Chapter 2 Lecture 3, Slide 1 Review: Power If an element is