3. Voltage and Current laws

Transcription

1 1 3. Voltage and Current laws 3.1 Node, Branches, and loops A branch represents a single element such as a voltage source or a resistor A node is the point of the connection between two or more elements (branches) It is usually indicated by a dot in a circuit If a connecting wire (short circuit) connects two nodes, the two nodes constitute a single nodes A loop is any closed path in a circuit A closed path is formed by starting at a node, passing through a set of nodes and returning to the start node without passing through any node more than once loop branch

2 2 3.2 Kirchhoff's Current Law Kirchhoff's Current Law (KCL) is based on the law of conservation of charge The algebraic sum of the currents entering any node is zero An alternative form of KCL is the current entering any node = the current leaving that node KCL can be applied to any closed boundary (closed path)

3 3 3.3 Kirchhoff's Voltage Law Kirchhoff's voltage Law (KVL) is based on the law of conservation of energy The algebraic sum of the voltages around any closed path is zero KVL When voltage sources are connected in series, KVL can be applied to obtain the total voltage a a b = b

4 4 Determine v x in the circuit Ans:

5 5 Determine v x in the circuit Ans:

6 6 3.4 The Single-Loop Circuit Single-loop circuits Elements are connected in series All elements carry the same current We shall determine The current through each element The voltage across each element The power absorbed by each element

7 7 3.4 The Single-Loop Circuit We apply the following steps 1) Assign a reference direction for the unknown current 2) Assign voltage references to the elements 3)Apply KVL to the closed loop path KVL 4)Use Ohm's law where needed to get an equation in i 5)Solve for i

8 8 Find i and p for all elements in the circuit Ans: KVL 1) Assign a reference direction for the unknown current 2)Assign voltage references to the elements (note that v A =-v 2 ) 3)Apply KVL to the closed loop path 4)Use Ohms law where needed to get an equation in i 5)Solve for i

9 9 Find i and p for all elements in the circuit Ans: KVL Computing the power absorbed by each element The total power absorbed by all elements

10 The Single-Node-Pair Circuit Single-node-pair circuits Elements are connected in parallel All elements have a common voltage We shall determine The current through each element The voltage across each element The power absorbed by each element

11 The Single-Node-Pair Circuit We apply the following steps 1) Define the voltage v and arbitrary select its polarity 2) Use passive sign convention to determine the currents directions 3)Apply KCL at the node 4)Use Ohm's law where needed to get an equation in v 5)Solve for v

12 12 Find v and p supplied by the independent source Ans: 1) Assign an arbitrary sign for the unknown voltage 2)passive sign convention to find the currents directions (note that ix=-i 2 ) 3)Apply KCL to the nodes 4)Use Ohm's law where needed to get an equation in v 5)Solve for v

13 13 HW: Find i 1, i 2, i 3, and i 4

14 Series and Parallel Connected Sources Series-connected voltage sources can be replaced by a single source Parallel current sources can be replaced by a single source

15 Resistors Series and Parallel Series connection KVL Parallel connection

16 16 Find the voltage and the power of the independent source 1)Apply KCL at the top node 2)Use Ohm's law for (i 1 =v x /6) and (v x =3i 3 ) 3) Solve i 3 and v x

17 Voltage and Current Division Voltage divider: is a passive linear circuit that produces an output voltage (v out ) that is a fraction of its input voltage (v in ) Easily solved with KCL, KVL, & equivalent resistances Then, Generally, assume we have The voltage v N can be given as Easy to find the other voltages, too

18 18

19 19 Current divider: is a simple linear circuit that produces an output current (i out ) that is a fraction of its input current (i in ) Easily solved with Since For n=2, we have The circuit divider reduces to

20 20 Use resistance combination methods and current division to find i 1 and i 2 and v x Ans: We note i 1 goes to the following equivalent resistor Use current divider, we have

21 21 We note i 2 goes to the following equivalent resistor Use current divider, we have HW: Solve v x

22 22 We note i 2 goes to the following equivalent resistor Use current divider, we have HW: Solve v x Homework Assignment 2 P3.6, P3.7, P3.13, P3.15, P3.16, P3.19, P3.20, P3.21, P3.30, P3.31, P3.35, P3.39, P3.73, P3.75 and P3.82