1. Lecture 4, Slide 1EE 40 Fall 2004Prof. White 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. Lecture 4, Slide 2EE 40 Fall 2004Prof. White 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 Equivalent resistance of resistors in series is the sum R2R2 R1R1 V SS I R3R3 R4R4  + Resistors in Series

3. Lecture 4, Slide 3EE 40 Fall 2004Prof. White I = V SS / (R 1 + R 2 + R 3 + R 4 ) Voltage Divider +–+– V1V1 +–+– V3V3 R2R2 R1R1 V SS I R3R3 R4R4  +

4. Lecture 4, Slide 4EE 40 Fall 2004Prof. White SS 4321 2 2 V RRRR R V    Correct, if nothing else is connected to nodes because R 5 removes condition of resistors in series SS 4321 2 2 V RRRR R V   ≠ When can the Voltage Divider Formula be Used? +–+– V2V2 R2R2 R1R1 V SS I R3R3 R4R4  + R2R2 R1R1 I R3R3 R4R4  + R5R5 +–+– V2V2

5. Lecture 4, Slide 5EE 40 Fall 2004Prof. White 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” ( R in ). Ideally, this should be very high (typical value 10 M  ) Ideal Voltmeter R in Measuring Voltage

6. Lecture 4, Slide 6EE 40 Fall 2004Prof. White Example: V SS R1R1 R2R2 If V SS R1R1 R2R2 R in Effect of Voltmeter undisturbed circuit circuit with voltmeter inserted _ + _ + +–+– V2V2 +–+– V2′V2′

7. Lecture 4, Slide 7EE 40 Fall 2004Prof. White KVL tells us that the same voltage is dropped across each resistor V x = I 1 R 1 = I 2 R 2 KCL tells us R2R2 R1R1 I SS I2I2 I1I1 x Resistors in Parallel Consider a circuit with two resistors connected in parallel. Find their “equivalent resistance”.

8. Lecture 4, Slide 8EE 40 Fall 2004Prof. White What single resistance R eq is equivalent to three resistors in parallel? +  V I V +  I R3R3 R2R2 R1R1 R eq eq  General Formula for Parallel Resistors Equivalent conductance of resistors in parallel is the sum

9. Lecture 4, Slide 9EE 40 Fall 2004Prof. White V x = I 1 R 1 = I SS R eq Current Divider R2R2 R1R1 I SS I2I2 I1I1 x

10. Lecture 4, Slide 10EE 40 Fall 2004Prof. White +  V Generalized Current Divider Formula Consider a current divider circuit with >2 resistors in parallel:

11. Lecture 4, Slide 11EE 40 Fall 2004Prof. White Find i 2, i 1 and i o Circuit w/ Dependent Source Example

12. Lecture 4, Slide 12EE 40 Fall 2004Prof. 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

13. Lecture 4, Slide 13EE 40 Fall 2004Prof. 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

14. Lecture 4, Slide 14EE 40 Fall 2004Prof. 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

15. Lecture 4, Slide 15EE 40 Fall 2004Prof. 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

16. Lecture 4, Slide 16EE 40 Fall 2004Prof. 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

17. Lecture 4, Slide 17EE 40 Fall 2004Prof. 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

18. Lecture 4, Slide 18EE 40 Fall 2004Prof. 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

19. Lecture 4, Slide 19EE 40 Fall 2004Prof. White V 2 V 1 R 2 R 1 R 4 R 5 R 3 I 1 VaVa Nodal Analysis: Example #2

20. Lecture 4, Slide 20EE 40 Fall 2004Prof. 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”

21. Lecture 4, Slide 21EE 40 Fall 2004Prof. 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