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

Basic Electrical Concepts
73
nonstandard resistance values. Another practical application of the principles
in this exercise is to distribute power dissipation to multiple components.
You will probably run into situations in which the power
dissipation requirements for a resistor will exceed its maximum specification.
In these cases, you can distribute the total power dissipation to
multiple resistors, thus decreasing the individual power dissipation
requirements. For example, if you inserted a single 700-ohm resistor in
place of the parallel network of R1 and R2 in Fig. 2-23a, the power dissipation
requirement on this single resistor would be
P IE (10 milliamps)(7 volts) 70 mW or 0.07 watts
By using two 1.4-Kohm resistors in parallel as originally illustrated in
Fig. 2-23a, the power dissipation requirement on “each” resistor would be
P IE (5 milliamps)(7 volts) 35 mW or 0.035 watts
Note how the “total” power dissipation remained at 70 milliwatts (with
two resistors dissipating 35 milliwatts, the total dissipation is 35 mW 35
mW 70 mW), but by incorporating the parallel network, each individual
resistor was required to dissipate only 35 milliwatts.
Exercise 11 Referring to Fig. 2-23b, what should the resistance value of
R1 be, assuming both R1 and R2 to have the same resistance value This
is another example of a series-parallel circuit. Note that D1 is in series
with R3, and R3 is in series with the parallel network of R1 and R2. The
total circuit current flow must flow through D1 and R3, but the current
branches at node 1, with part of the current flowing through the
R1 leg and the remainder of it flowing through the R2 leg. The two
currents sum at node 2, with the total circuit current flowing to the
positive terminal of the source.
The first step required in this exercise is to determine the voltage
drop across R3. Since R3 and D1 are in series, and 10 milliamps should
flow through D1, you know that 10 milliamps must flow through R3 as
well. Now that you know two electrical variables specifically relating to
R3 (i.e., its resistance value and the current flow through it), you can calculate
the voltage drop across it using Eq. (2-1):
E IR (10 milliamps)(100 ohms) 1 volt
Since D1 is dropping 2 volts and R3 is dropping 1 volt, the remainder
of the source voltage must be dropped across the parallel network of