1. EECS 42 fall 2004 lecture 5 9/10/2004 EE 42 lecture 5 Equivalent resistance of a passive linear circuit Thevenin’s Theorem for a single source linear circuit Norton’s Theorem for a single source linear circuit Measurements Nodal analysis

2. EECS 42 fall 2004 lecture 5 9/10/2004 Linear equations-> straight line We know it is a straight line because the circuit gives us a set of linear equations, and so they have a linear solution. V I +V-+V- I

3. EECS 42 fall 2004 lecture 5 9/10/2004 Equivalent resistance for a linear passive circuit. If we have a circuit which includes only wires and resistors, the relationship between the voltage and current are related by a straight line, through the origin. V I +V-+V- I

4. EECS 42 fall 2004 lecture 5 9/10/2004 Why it must pass through the origin if there are no sources of power The function of current vs voltage must pass through the origin, otherwise it would pass through a quadrant where we could extract power—and we said there were no sources of power! V I +V-+V- I

5. EECS 42 fall 2004 lecture 5 9/10/2004 Circuits with sources If we have any circuit which includes DC sources (voltage or current) and linear resistances, then the I-V curve is still linear, but it does not have to pass through the origin. V I I +V-+V-

6. EECS 42 fall 2004 lecture 5 9/10/2004 Models and measurements of circuits with sources Since the I-V curve is a straight line, we can characterize it with two parameters, for example:  The current intercept and slope.  The voltage intercept and slope.  Both the current and voltage intercept. If we measure any two points, we can calculate any of these parameters.

7. EECS 42 fall 2004 lecture 5 9/10/2004 Thevenin’s Theorm We can make an equivalent circuit for any circuit which includes only resistors and linear sources with a single voltage source and a resistor. This translates the voltage intercept and slope equation into a circuit. V I I +V-+V- ~

8. EECS 42 fall 2004 lecture 5 9/10/2004 Thevenin’s Theorm The point where the current is equal to zero gives us the voltage put out by the voltage source in the equivalent circuit. (1/slope) gives us the resistance R of the resistor in the equivalent circuit. V I I +V-+V- ~ +V-+V- R

9. EECS 42 fall 2004 lecture 5 9/10/2004 Thevenin’s Theorm A large slope means a small value of R V I I +V-+V- ~ +V-+V- R

10. EECS 42 fall 2004 lecture 5 9/10/2004 Norton’s Theorem We can also make a different equivalent circuit, for any circuit which contains only linear sources and resistors, with a current source The needed value of current from the current source is the amount of current put out by the actual circuit when the voltage is zero (the current intercept) This model is called a Norton equivalent circuit.

11. EECS 42 fall 2004 lecture 5 9/10/2004 Norton’s Theorem The point where the voltage is equal to zero gives us the current put out by the voltage source in the equivalent circuit. (1/slope) gives us the resistance R of the resistor in the equivalent circuit. V I I +V-+V- R I

12. EECS 42 fall 2004 lecture 5 9/10/2004 Taking Measurements To measure voltage, we use a two-terminal device called a voltmeter. To measure current, we use a two-terminal device called a ammeter. To measure resistance, we use a two-terminal device called a ohmmeter. A multimeter can be setup to function as any of these three devices. In lab, you use a DMM to take measurements, which is short for digital multimeter.

13. EECS 42 fall 2004 lecture 5 9/10/2004 Measuring Current To measure current, insert the measuring instrument in series with the device you are measuring. That is, put your measuring instrument in the path of the current flow. The measuring device will contribute a very small resistance (like wire) when used as an ammeter. It usually does not introduce serious error into your measurement, unless the circuit resistance is small. i DMM

14. EECS 42 fall 2004 lecture 5 9/10/2004 Measuring Voltage To measure voltage, insert the measuring instrument in parallel with the device you are measuring. That is, put your measuring instrument across the measured voltage. The measuring device will contribute a very large resistance (like air) when used as a voltmeter. It usually does not introduce serious error into your measurement unless the circuit resistance is large. + v - DMM

15. EECS 42 fall 2004 lecture 5 9/10/2004 Example For the above circuit, what is i 1 ? Suppose i 1 was measured using an ammeter with internal resistance 1 Ω. What would the meter read? 9 Ω 27 Ω i1i1 i2i2 i3i3 54 Ω 3 A

16. EECS 42 fall 2004 lecture 5 9/10/2004 Measuring Resistance To measure resistance, insert the measuring instrument in parallel with the resistor you are measuring with nothing else attached. The measuring device applies a voltage to the resistance and measures the current, then uses Ohm’s law to determine resistance. It is important to adjust the settings of the meter for the approximate size (Ω or MΩ) of the resistance being measured so appropriate voltage is applied to get a reasonable current. DMM

17. EECS 42 fall 2004 lecture 5 9/10/2004 Measurements to derive a Thevenin equivalent circuit Measure the open circuit voltage. The voltage for the voltage source in the Tevenin model is this open circuit voltage. Measure the short circuit current. The resistance to put into the Thevenin model is R=Voc/Iss Alternatively, you can turn off all of the sources within the black box and measure its resistance  Turn off -> voltage sources to 0 volts (replace them with a short circuit), current sources to zero current (open)

18. EECS 42 fall 2004 lecture 5 9/10/2004 Measurements to derive a Norton equivalent circuit Measure the short circuit current The current for the current source in the Norton model is this open circuit voltage. Measure the open circuit voltage. The resistance to put into the Norton model is R=Voc/Iss Alternatively, you can turn off all of the sources within the black box and measure its resistance  Turn off -> voltage sources to 0 volts (replace them with a short circuit), current sources to zero current (open)

19. EECS 42 fall 2004 lecture 5 9/10/2004 Solution to Example: By current division, i 1 = -3 A (18 Ω)/(9 Ω+18 Ω) = -2 A When the ammeter is placed in series with the 9 Ω, Now, i 1 = -3 A (18 Ω)/(10 Ω+18 Ω) = -1.93 A 9 Ω 27 Ω i1i1 i2i2 i3i3 54 Ω 3 A 9 Ω 18 Ω i1i1 3 A 9 Ω 27 Ω i1i1 i2i2 i3i3 54 Ω 3 A 10 Ω 18 Ω i1i1 3 A 1 Ω