Transcript ch2.10_applicationex..

V2
P
 SMALLER RESISTANCE FOR HIGHER POWER
R
A : R   (OFF )
B : R  R2
LOW
C : R  R1
MEDIUM
D : R  R1 || R2
PMEDIUM
PHIGH
230[V ]2
 2000W 
 R1 || R2  26.45
R1 || R2
R1 R2
 26.45
R1  R2
OR
HIGH
230[V ]2
 1200W 
 R1  44.08
R1
230[V ]2
PR  1200W  PR  800W 
R2
1
2
 1A
This is the set up in the car
 100A
Circuit Representation
WHEN THE STARTER IS ENGAGED THE VOLTAGE DROP ACROSS THE RESISTENCE
INCREASES 100 FOLD --- LIGHTS DIM!
Measure R3 with scale empty
Measure R3 when truck is on sclae
Weight is determined in the following manner.
The change in R3 required to balance the bridge
represents the change in strain, which when
RX 
R2
R
R1 3
Balance condition
For bridge circuit
multiplied by the modulus of elasticity yields
the change in stress.
The change in stress multiplied by the
cross-sectional area of the cylinder produces
the change in load, which determines the weight
APPPLICATIONS
http://www.wiley.com/college/irwin/0470128690/animations/swf/2-34.swf
Spec: V2  9V  5%  8.55V  V2  9.45V
rating: 9V / 5mA 1.8k
Highest V2: No lamp connected
R1
R2

 0.27
9.45 
 12
R2
R1  R2
Lowest V2: All lamps ON
(5 lamps in parallel with R2)
1
1
1
5

 ... 

 RP 5  360
RP 5 1800
1800 1800
Equivalent circuit
8.45 
R2 || 360
 12
R1  ( R2 || 360)
360 R1
 360  R1
R2
12
R

360
8.55
1
 48.1
Design a current-to-voltage converter that will output 5 V when the current signal
Is 20mA
Assume that the controller does not
use any current from the I to V converter.
In this case a simple resistor behaves as
current to voltage converter!
Wire resistance does not matter!
(within reasonable limits)
R
Design Spec: R  20mA  5V
More realistic model – I_signal is NOT I_sensor.
There is a current division determined by Rs and R_wire
(length of wire becomes important)
 R  250
DESIGNS