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Week2bEECS 42, Spring 2005Prof. White Find i 2, i 1 and i o Circuit w/ Dependent Source Example

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Week2bEECS 42, Spring 2005Prof. White Simplify a circuit before applying KCL and/or KVL: + 7 V Using Equivalent Resistances R 1 = R 2 = 3 k R 3 = 6 k R 4 = R 5 = 5 k R 6 = 10 k I R1R1 R2R2 R4R4 R5R5 R3R3 R6R6 Example: Find I

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Week2bEECS 42, Spring 2005Prof. White The Wheatstone Bridge Circuit used to precisely measure resistances in the range from 1 to 1 M , with ±0.1% accuracy R 1 and R 2 are resistors with known values R 3 is a variable resistor (typically 1 to 11,000 ) R x is the resistor whose value is to be measured +V–+V– R1R1 R2R2 R3R3 RxRx current detector battery variable resistor

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Week2bEECS 42, Spring 2005Prof. White Finding the value of R x Adjust R 3 until there is no current in the detector Then, +V–+V– R1R1 R2R2 R3R3 RxRx R x = R 3 R2R1R2R1 Derivation: i 1 = i 3 and i 2 = i x i 3 R 3 = i x R x and i 1 R 1 = i 2 R 2 i 1 R 3 = i 2 R x KCL => KVL => R3R1R3R1 RxR2RxR2 = i1i1 i2i2 ixix i3i3 Typically, R 2 / R 1 can be varied from 0.001 to 1000 in decimal steps

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Week2bEECS 42, Spring 2005Prof. White Some circuits must be analyzed (not amenable to simple inspection) Special cases: R 3 = 0 OR R 3 = R1R1 + R4R4 R5R5 R2R2 V R3R3 Identifying Series and Parallel Combinations

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Week2bEECS 42, Spring 2005Prof. White Resistive Circuits: Summary Equivalent resistance of k resistors in series: R eq = R i = R 1 + R 2 + + R k Equivalent resistance of k resistors in parallel: Voltage divided between 2 series resistors: Current divided between 2 parallel resistors: i=1 k 1 R eq = + i=1 k 1Ri1Ri 1R11R1 1R21R2 1Rk1Rk v 1 = v s R 1 R 1 + R 2 i 1 = i s R 2 R 1 + R 2

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Week2bEECS 42, Spring 2005Prof. White Circuit Analysis Methods … Given: a circuit with voltage and/of current sources, and components (Rs, Cs, Ls, diodes, transistors, etc.) connected by wires Task: find all the voltages and currents of interest Approaches: 1.Use Ohm’s Law, KCL and KVL piecemeal 2.Nodal analysis – node voltages are the unknowns 3.Mesh (loop) analysis – mesh currents are the unknowns 4.Superposition (linear circuits) – work with only one voltage or current source at a time) and use #1, #2 or #3 5.Computer simulation of circuit behavior

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Week2bEECS 42, Spring 2005Prof. White 1.Choose a reference node (“ground”) Look for the one with the most connections! 2.Define unknown node voltages those which are not fixed by voltage sources 3.Write KCL at each unknown node, expressing current in terms of the node voltages (using the I-V relationships of branch elements) Special cases: floating voltage sources 4.Solve the set of independent equations N equations for N unknown node voltages Node-Voltage Circuit Analysis Method

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Week2bEECS 42, Spring 2005Prof. White 1.Choose a reference node. 2.Define the node voltages (except reference node and the one set by the voltage source). 3.Apply KCL at the nodes with unknown voltage. 4.Solve for unknown node voltages. R 4 V 1 R 2 + - ISIS R 3 R1R1 Nodal Analysis: Example #1

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Week2bEECS 42, Spring 2005Prof. White

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Week2bEECS 42, Spring 2005Prof. White V 2 V 1 R 2 R 1 R 4 R 5 R 3 I 1 VaVa Nodal Analysis: Example #2

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Week2bEECS 42, Spring 2005Prof. White

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Week2bEECS 42, Spring 2005Prof. White A “floating” voltage source is one for which neither side is connected to the reference node, e.g. V LL in the circuit below: Problem: We cannot write KCL at nodes a or b because there is no way to express the current through the voltage source in terms of V a -V b. Solution: Define a “supernode” – that chunk of the circuit containing nodes a and b. Express KCL for this supernode. Incorporate voltage source constraint into KCL equation. R 4 R 2 I 2 V a V b + - V LL I 1 Nodal Analysis w/ “Floating Voltage Source”

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Week2bEECS 42, Spring 2005Prof. White supernode Eq’n 1: KCL at supernode Substitute property of voltage source: R 4 R 2 I 2 V a V b + - V LL I 1 Nodal Analysis: Example #3

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Week2bEECS 42, Spring 2005Prof. White

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Week2bEECS 42, Spring 2005Prof. White Node-Voltage Method and Dependent Sources If a circuit contains dependent sources, what to do? Example: –+–+ –+–+ 80 V 5i5i 20 10 200 2.4 A ii

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Week2bEECS 42, Spring 2005Prof. White Node-Voltage Method and Dependent Sources Dependent current source: treat as independent current source in organizing and writing node eqns, but include (substitute) constraining dependency in terms of defined node voltages. Dependent voltage source: treat as independent voltage source in organizing and writing node eqns, but include (substitute) constraining dependency in terms of defined node voltages.

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Week2bEECS 42, Spring 2005Prof. White –+–+ –+–+ 80 V 5i5i 20 10 200 2.4 A ii Example:

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