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Lecture #4

Lecture #4. OUTLINE Resistors in series equivalent resistance voltage-divider circuit measuring current Resistors in parallel equivalent resistance current-divider circuit measuring voltage Examples Node Analysis Reading Chapter 2. +. . Resistors in Series.

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Lecture #4

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  1. Lecture #4 OUTLINE • Resistors in series • equivalent resistance • voltage-divider circuit • measuring current • Resistors in parallel • equivalent resistance • current-divider circuit • measuring voltage • Examples • Node Analysis Reading Chapter 2

  2. +  Resistors in Series Consider a circuit with multiple resistors connected in series. Find their “equivalent resistance”. • KCL tells us that the same current (I) flows through every resistor • KVL tells us I R1 R2 VSS R3 R4 Equivalent resistance of resistors in series is the sum

  3. +  Voltage Divider I = VSS / (R1 + R2 + R3 + R4) I + – V1 R1 R2 + – VSS V3 R3 R4

  4. + +   R R 2 2 V V × × V V = ≠ 2 2 SS SS R R + + R R + + R R + + R R 1 1 2 2 3 3 4 4 When can the Voltage Divider Formula be Used? I I R1 R1 + – + – R2 R2 V2 V2 VSS VSS R3 R3 R4 R4 R5 Correct, if nothing else is connected to nodes because R5 removes condition ofresistors in series

  5. Measuring Voltage To measure the voltage drop across an element in a real circuit, insert a voltmeter (digital multimeter in voltage mode) in parallel with the element. Voltmeters are characterized by their “voltmeter input resistance” (Rin). Ideally, this should be very high (typical value 10 MW) Ideal Voltmeter Rin

  6. Effect of Voltmeter circuit with voltmeter inserted undisturbed circuit R1 R1 + – + – _ + _ + VSS V2 VSS V2′ R2 R2 Rin Example: If

  7. Resistors in Parallel Consider a circuit with two resistors connected in parallel. Find their “equivalent resistance”. x • KVL tells us that the • same voltage is dropped • across each resistor • Vx = I1 R1 = I2 R2 • KCL tells us I1 I2 ISS R1 R2

  8. + + I I   V eq  R1 R2 R3 V Req General Formula for Parallel Resistors What single resistance Req is equivalent to three resistors in parallel? Equivalent conductance of resistors in parallel is the sum

  9. Current Divider x I1 I2 Vx = I1 R1 = ISS Req ISS R1 R2

  10. + V  Generalized Current Divider Formula Consider a current divider circuit with >2 resistors in parallel:

  11. Circuit w/ Dependent Source Example Find i2, i1 and io

  12. +  Using Equivalent Resistances Simplify a circuit before applying KCL and/or KVL: Example: Find I I R1 R1 = R2 = 3 kW R3 = 6 kW R3 R2 7 V R4 R4 = R5 = 5 kW R6 = 10 kW R6 R5

  13. The Wheatstone Bridge • Circuit used to precisely measure resistances in the range from 1 W to 1 MW, with ±0.1% accuracy • R1and R2 are resistors with known values • R3 is a variable resistor (typically 1 to 11,000W) • Rx is the resistor whose value is to be measured battery R1 R2 + V – current detector R3 Rx variable resistor

  14. Finding the value of Rx • Adjust R3 until there is no current in the detector Then, R2 R1 Rx = R3 • Derivation: • i1 = i3 and i2 = ix • i3R3 = ixRx and i1R1 = i2R2 • i1R3 = i2Rx KCL => KVL => R1 R2 i1 i2 + V – i3 ix R3 Rx R3 R1 Rx R2 = Typically, R2 / R1 can be varied from 0.001 to 1000 in decimal steps

  15. R2 R1 R3 V R5 R4 +  Identifying Series and Parallel Combinations Some circuits must be analyzed (not amenable to simple inspection) Special cases: R3 = 0 OR R3 = 

  16. Resistive Circuits: Summary • Equivalent resistance of kresistorsin series: Req = SRi = R1 + R2 + ••• + Rk • Equivalent resistance of k resistorsin parallel: • Voltage divided between 2 series resistors: • Current divided between 2 parallel resistors: k i=1 1 Req 1 Ri 1 Rk 1 R1 1 R2 k = S = + + ••• + i=1 R1 R1 + R2 v1 = vs R2 R1 + R2 i1 = is

  17. Node-Voltage Circuit Analysis Method • Choose a reference node (“ground”) Look for the one with the most connections! • Define unknown node voltages those which are not fixed by voltage sources • 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 • Solve the set of independent equations N equations for N unknown node voltages

  18. R1 R 3 + IS V1 R R - 2 4 Nodal Analysis: Example #1 • Choose a reference node. • Define the node voltages (except reference node and the one set by the voltage source). • Apply KCL at the nodes with unknown voltage. • Solve for unknown node voltages.

  19. R 1 R 5 R I 3 1 V V R R 2 1 2 4 Nodal Analysis: Example #2 Va

  20. V V LL V a b - + I R I R 1 2 2 4 Nodal Analysis w/ “Floating Voltage Source” A “floating” voltage source is one for which neither side is connected to the reference node, e.g. VLL 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 Va-Vb. 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.

  21. Nodal Analysis: Example #3 supernode V V LL V a b - + I R I R 1 2 2 4 Eq’n 1: KCL at supernode Substitute property of voltage source:

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