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Announcements First Assignment posted: –Due in class in one week (Thursday Sept 15 th )

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Presentation on theme: "Announcements First Assignment posted: –Due in class in one week (Thursday Sept 15 th )"— Presentation transcript:

1 Announcements First Assignment posted: –Due in class in one week (Thursday Sept 15 th ) http://www.physics.udel.edu/~jholder/Phys645/index.htm

2 Lecture 3 Overview Loop Analysis with KVL & KCL Mesh Analysis Thevenin/Norton equivalent circuits

3 Circuit analysis method 2a: KVL and KCL Kirchoff’s Voltage Law: Loop analysis The sum of the voltages around a closed loop must be zero Draw the current direction in every branch (arbitrary) and label the voltage directions (determined by the defined current direction). Voltage on a voltage source is always from positive to negative end. Define either clockwise or counter- clockwise as positive direction for summing voltages. Once the direction is defined, use the same convention in every loop. Voltage across a resistor is +’ve if voltage direction the same as current direction, -’ve otherwise Apply KVL

4 Kirchoff’s Voltage Law: Multiloop The sum of the voltages around a closed loop must be zero Draw the current direction (arbitrary) and label the voltage directions (determined by the defined current direction). Define either clockwise or counter- clockwise as positive voltage direction. Once the direction is defined, use the same convention in every loop. Apply KVL R3R3 Say r=1Ω, R 1 =3Ω, R 2 =5Ω, R 3 =10Ω, ε=3V What are the currents?

5 Kirchoff’s Current Law The sum of the current at a node must be zero: I in =I out R3R3 I=I 2 +I 3 (1) ε=Ir+IR 1 +I 2 R 2 (2) I 3 R 3 -I 2 R 2 =0 (3) 1I- 1I 2 - 1I 3 = 0 (4) 4I+5I 2 + 0I 3 = 3 (5) Say r=1Ω, R 1 =3Ω, R 2 =5Ω, R 3 =10Ω, ε=3V 0I- 5I 2 +10I 3 = 0 (6)

6 Last note on KCL KVL analysis If solutions to currents or voltages are negative, this just means the real direction is opposite to what you originally defined To deal with current sources: current is known, but assign a voltage across it which has to be solved

7 Another Sample Problem: Multiple Sources

8 Method 2b: Mesh Analysis Example: 2 meshes (Mesh is a loop that does not contain other loops) Step 1: Assign mesh currents clockwise Step 2: Apply KVL to each mesh The self-resistance is the effective resistance of the resistors in series within a mesh. The mutual resistance is the resistance that the mesh has in common with the neighbouring mesh To write the mesh equation, evaluate the self-resistance, then multiply by the mesh current Next, subtract the mutual resistance multiplied by the current in the neighbouring mesh for each neighbour. Equate the above result to the driving voltage: taken to be positive if it tends to push current in the same direction as the assigned mesh current Mesh1: (R 1 +R 2 )I 1 - R 2 I 2 =ε 1 -ε 2 Mesh2: -R 2 I 1 + (R 2 +R 3 )I 2 =ε 2 -ε 3 Step 3: solve currents

9 Method 2b: Mesh Analysis

10 Another mesh analysis example Find the currents in each branch Step 1: Replace any combination of resistors in series or parallel with their equivalent resistance Step 2: Assign mesh currents clockwise Step 3: Write the mesh equations for each mesh Left mesh: 11I 1 -6I 2 =9 Right mesh: - 6I 1 +18I 2 =9 Note:suppressed “k” for each resistor, so answer is in mA Step 4: Solve the equations Solution: I 1 =4/3mA =1.33mA I 2 =17/18mA =0.94mA

11 Mesh analysis with a current source Magnitude of current in branch containing current source is I S, (although if the current flow is opposite to the assigned current direction the value will be negative). This works only if the current source is not shared by any other mesh For a shared current source, label it with an unknown voltage.

12 Mesh analysis with mixed sources Find I x Identify mesh currents and label accordingly Write the mesh equations Mesh1: I 1 =-2 Mesh2:-4 I 1 +8I 2 -4I 4 =12 Mesh3: 8I 3 =-12 Mesh4:-2 I 1 -4I 2 +6I 4 =10 I 1 = -2.0A I 2 = 1.5A I 3 = -1.5A I 4 = 2.0A I x =I 2 -I 3 I x =3.0A

13 Method 3: Thevenin and Norton Equivalent Circuits v TH = open circuit voltage at terminal (a.k.a. port) R TH = Resistance of the network as seen from port (V m ’s, I n ’s set to zero) Any network of sources and resistors will appear to the circuit connected to it as a voltage source and a series resistance

14 Norton Equivalent Circuit Any network of sources and resistors will appear to the circuit connected to it as a current source and a parallel resistance

15 Calculation of R T and R N R T =R N ; same calculation (voltage and current sources set to zero) Remove the load. Set all sources to zero (‘kill’ the sources) –Short voltage sources (replace with a wire) –Open current sources (replace with a break)

16 Calculation of R T and R N continued Calculate equivalent resistance seen by the load

17 Calculation of V T Remove the load and calculate the open circuit voltage

18 Calculation of I N Short the load and calculate the short circuit current (R 1 +R 2 )i 1 - R 2 i SC = v s -R 2 i 1 + (R 2 +R 3 )i SC = 0 KCL

19 Source Transformation Summary: Thevenin’s Theorem Any two-terminal linear circuit can be replaced with a voltage source and a series resistor which will produce the same effects at the terminals V TH is the open-circuit voltage V OC between the two terminals of the circuit that the Thevenin generator is replacing R TH is the ratio of V OC to the short-circuit current I SC ; In linear circuits this is equivalent to “killing” the sources and evaluating the resistance between the terminals. Voltage sources are killed by shorting them, current sources are killed by opening them.

20 Summary: Norton’s Theorem Any two-terminal linear circuit can be replaced with a current source and a parallel resistor which will produce the same effects at the terminals I N is the short-circuit current I SC of the circuit that the Norton generator is replacing Again, R N is the ratio of V OC to the short-circuit current I SC ; In linear circuits this is equivalent to “killing” the sources and evaluating the resistance between the terminals. Voltage sources are killed by shorting them, current sources are killed by opening them. For a given circuit, R N =R TH

21 Maximum Power Transfer Why use Thevenin and Norton equivalents? –Very easy to calculate load related quantities –E.g. Maximum power transfer to the load It is often important to design circuits that transfer power from a source to a load. There are two basic types of power transfer –Efficient power transfer (e.g. power utility) –Maximum power transfer (e.g. communications circuits) Want to transfer an electrical signal (data, information etc.) from the source to a destination with the most power reaching the destination. There is limited power at the source and power is small so efficiency is not a concern.

22 Maximum Power Transfer: Impedance matching so maximum power transfer occurs whenand Differentiate using quotient rule: Set to zero to find maximum: http://circuitscan.homestead.com/files/ancircp/maxpower1.htm


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