Tab Electronics Guide to Understanding Electricity ... - Sciences Club

74 Chapter Two
R1 and R2. Therefore, the voltage across R1 and R2 is 6 volts. Remembering
that R1 and R2 are assumed to have equal resistance values, you
know that the total circuit current will divide evenly at node 1, meaning
that 5 milliamps will flow through R1 and the remaining 5 milliamps
will flow through R2. Now you have two electrical variables
relating specifically to R1, its current flow (5 milliamps) and the voltage
across it (6 volts), using Eq. (2-3), its resistance value will be
E 6 volts
R
1200 ohms or 1.2 Kohm
I 5 milliamps
What is the practical value of this exercise In Exercise 10, you examined
how it was possible to use two unequal resistance values connected
in parallel to come up with unusual, or “nonstandard” resistance values.
By connecting a 1.2-Kohm resistor and a 1.8-Kohm resistor in parallel, it
was possible to derive an R equiv
resistance of 720 ohms, which was relatively
close to the desired target resistance of 700 ohms. This exercise shows
how it is possible to use a series-parallel resistor network to obtain
“exactly” 700 ohms. If R1 and R2 are each 1.2 Kohms, the R equiv
is 600
ohms. Adding this to the value of R3 (100 ohms) provides a total resistance
of 700 ohms. Also, the resistance values of 1.2 Kohms and 100 ohms
are standard resistor values. Using various combinations of series-parallel
circuits, it is possible to obtain virtually any resistance value for virtually
any application.
Exercise 12 Referring to Fig. 2-23c, what would be the required resistance
value of R1 The first step in solving a problem of this sort is to be
certain that you have properly conceptualized the circuit. Note that R2 and
R3 are in series, with this series network in parallel with R1, and this
entire series-parallel network in series with R4, which is in series with D1.
If this is confusing, compare Fig. 2-23c to Fig. 2-23d. They are electrically
identical, but Fig. 2-23d is redrawn to make it easier to visualize.
Since D1 is in series with R4, remembering that 10 milliamps of current
should be flowing through D1, you know that 10 milliamps should
also flow through R4. With this information, you can calculate the voltage
drop across R4 with Eq. (2-1):
E IR (10 milliamps)(100 ohms) 1 volt
Since D1 is dropping 2 volts and R4 is dropping 1 volt, the remaining
source voltage must be dropped across the series-parallel network of R1,
R2, and R3. In other words, D1 and R4 are dropping a total of 3 volts, so