Agenda Voltage division Current division Wheatstone bridge
Voltage Divider Circuit Developing more than one voltage level from a single voltage supply
Identify the current and Apply KVL
Connect a “Load” Resistor in Parallel
Determine v o
The Voltage Divider Rule Voltage dropped across each resistor may be determined by the voltage across any other resistor (or combination of resistors) by using the voltage divider rule expressed as: The subscripts must match (x and y)
Voltage Divider Rule Application If a single resistor is very large compared to the other series resistors, the voltage across that resistor will be the source voltage If the resistor is very small, the voltage across it will be essentially zero
Voltage Divider Rule Application If a resistor is more than 100 times larger than another resistor ▫Smaller resistor can be neglected
Voltage Division Voltage drop across a single resistor is proportional to the voltage drop across series connected resistance Constant of proportionality is ratio of specific resistance to equivalent resistance.
Current division Current at specific resistance is proportional to the total current applied to set of parallel connected resistance. Constant of proportionality is ratio of equivalent resistance to parallel resistance.
3 bulb question The circuit above shows three identical light bulbs attached to an ideal battery. If the bulb#2 burns out, which of the following will occur? a)Bulbs 1 and 3 are unaffected. The total light emitted by the circuit decreases. b)Bulbs 1 and 3 get brighter. The total light emitted by the circuit is unchanged. c)Bulbs 1 and 3 get dimmer. The total light emitted by the circuit decreases. d)Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit is unchanged. e)Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit is unchanged. f)Bulb 1 gets dimmer, but bulb 3 gets brighter. The total light emitted by the circuit decreases. g)Bulb 1 gets brighter, but bulb 3 gets dimmer. The total light emitted by the circuit decreases. h)Bulb 1 is unaffected, but bulb 3 gets brighter. The total light emitted by the circuit increases. i)None of the above.
When the bulb #2 is not burnt out: For Bulb #1 For Bulb #2 For Bulb #3
For Bulb #1 For Bulb #2 For Bulb #3 So, Bulb #1 gets dimmer and bulb #3 gets brighter. And the total power decreases. f) is the answer. Before total power was After total power is When the bulb #2 is burnt out:
Classification of Resistance Low resistance is the range of.1 to 1 ohm. Medium resistance is the range of 1 to low megha ohm. High resistance is.1M to higher range.
Wheatstone Bridge An Overview
History The Wheatstone Bridge was invented in 1833 by Samuel Hunter Christie Later named after Sir Charles Wheatstone for his many applications of the circuit through the 1840s The most common procedure for the bridge remains the testing of unknown electrical resistance
How Does it Work? Uses ratio of 3 known resistors Measures fourth unknown resistance
How Does it Work? (cont.) By changing resistors to adjusting variable resistors to balance the device, the mathematical ratio is used to calculate the fourth (unknown) resistance
Impact of the Wheatstone Bridge The Wheatstone Bridge is a very simple design, although there are more complex versions of achieving the same outcome Can be adjusted easily Fairly inexpensive to produce Also indirectly measures any variable that would change the resistance of a material ▫Ex: temperature, force, pressure
Little about Instruments……….. Galvanometers:a coil in a magnetic field that senses current. Ammeters:measures current. Voltmeter:measures voltage. Ohmmeters:measures resistance. Multimeters:one device that does all the above. Galvanometer is a needle mounted to a coil that rotates in a magnetic field. The amount of rotation is proportional to the current that flows through the coil. Symbolically we write Usually when
Analysis Identify the currents
Consider the bridge at “balance”, i g =0
Use to Measure Resistance Ratio Arms
Review of few last slides Created in 1833, popularized in 1840s Wheatstone bridges are one of the best methods of measuring resistance due to the basic mathematical ratio involved. Accurate standards with sensitive enough voltmeter, measurements of resistance within.05% can be reached. Many calibration laboratories still use this method today. The Wheatstone Bridge are replaceable; however, for its simplicity and versatility the circuit is an indispensible piece of technology
Wye – Delta Transformations Also known as T- transformations
Delta-to-Wye Transformations Delta-to-wye, or wye-to-delta are also sometimes called pi-to-tee or tee-to-pi transformations. These are equivalent circuit pairs. They apply for parts of circuits that have three terminals. Each version of the equivalent circuit has three resistors.
WHY Wye to Delta Transformation????????? We cannot use resistors in parallel. We cannot use resistors in series. If we knew V and I we could solve R eq = V I There is another way to solve the problem without solving for I (given, assume, V) and calculating R eq for V/I.
Delta-to-Wye Transformations Three resistors in a part of a circuit with three terminals can be replaced with another version, also with three resistors. The two versions are shown here. Note that none of these resistors is in series with any other resistor, nor in parallel with any other resistor. The three terminals in this example are labeled A, B, and C.
Notes on Names When we go from the delta connection (on the left) to the wye connection (on the right), we call this the delta-to-wye transformation. Going in the other direction is called the wye-to-delta transformation. One can go in either direction, as needed. These are equivalent circuits.
Each resistor in the Y network is the product of the resistors in the two adjacent ∆ branches, divided by the sum of the three ∆ resistors. Each resistor in the ∆ network is the sum of all possible products of Y resistors taken two at a time, divided by the opposite Y resistor.
Delta-to-Wye Transformation Equations When we perform the delta-to-wye transformation (going from left to right) we use the equations given below.
Wye-to-Delta Transformation Equations When we perform the wye-to-delta transformation (going from right to left) we use the equations given below.
Why Are Delta-to-Wye Transformations Needed? This is a good question. In fact, it should be pointed out that these transformations are not necessary. Rather, they are like many other aspects of circuit analysis in that they allow us to solve circuits more quickly and more easily. They are used in cases where the resistors are neither in series nor parallel, so to simplify the circuit requires something more. One key in applying these equivalents is to get the proper resistors in the proper place in the equivalents and equations. We recommend that you name the terminals each time, on the circuit diagrams, to help you get these things in the right places.
Simplification If R1 = R2 = R3 = R, then Ra = Rb =Rc = 3R If Ra = Rb = Rc = R’, then R1 = R2 = R3 = R’/3
Readings Chapter 3: 3.4, 3.5, 3.6, 3.7 (Electric Circuits) ▫By James W. Nilson
Quiz 1. The total resistance of parallel resistors is equal to a. the sum of the resistances b. the sum of the reciprocals of the resistances c. the sum of the conductances d. none of the above
Quiz 2. The number of nodes in a parallel circuit is a. one b. two c. three d. can be any number
Quiz 3. The total resistance of the parallel resistors is a. 2.52 k b. 3.35 k c. 5.1 k d. 25.1 k
Quiz 4. If three equal resistors are in parallel, the total resistance is a. one third the value of one resistor b. the same as one resistor c. three times the value of one resistor d. there is not enough information to say
Quiz 5. In any circuit the total current entering a junction is a. less than the total current leaving the junction b. equal to the total current leaving the junction c. greater than the total current leaving the junction d. can be any of the above, depending on the circuit
Quiz 6. The current divider formula to find I 1 for the special case of two resistors is a. b. c. d.
Quiz 7. The total current leaving the source is a. 1.0 mA b. 1.2 mA c. 6.0 mA d. 7.2 mA
Quiz 8. The current in R 1 is a. 6.7 mA b. 13.3 mA c. 20 mA d. 26.7 mA
Quiz 9. The voltage across R 2 is a. 0 V b. 0.67 V c. 1.33 V d. 4.0 V
Quiz 10. The total power dissipated in a parallel circuit is equal to the a. power in the largest resistor b. power in the smallest resistor c. average of the power in all resistors d. sum of the power in all resistors
Quiz Answers: 1. d 2. b 3. a 4. a 5. b 6. c 7. d 8. b 9. c 10. d