BASIC ELECTRONICS - OD1633 - LESSON 1/TASK
It is apparent that if there are 5 amperes of current through each of the
two resistors, there must be a total current of 10 amperes drawn from the
source.
The total current of 10 amperes, as illustrated in figure 45, view B on the
previous page, leaves the negative terminal of the battery and flows to
point a. Since point a is a connecting point for the two resistors, it is
palled a junction.
At junction a, the total current divides into two
currents of 5 amperes each. These two currents flow through their resistors
and rejoin at junction b. The total current then flows from junction b back
to the positive terminal of the source. The source supplies a total current
of 10 amperes and each of the two equal resistors carries one-half of the
total current.
Each individual current path in the circuit of figure 45 (view B) is
referred to as a branch. Each branch carries a current that is a portion of
the total current. Two or more branches form a network.
From the previous explanation, the characteristics of current in a parallel
circuit can be expressed in terms of the following general equation:
IT = I1 + I2 + I3 + ... In
Compare view A of figure 46 (on the following page) with view B of the
circuit in figure 45. Notice that doubling the value of the second branch
resistor (R2) has no effect on the current in the first branch (IR1), but
does reduce the second branch current (IR2) to one-half its original value.
The total circuit current drops to a value equal to the sum of the branch
currents. These facts are verified by the following equations.
Given:
Es = 50 volts
R1 = 10 Ohms
R2 = 20 Ohms
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