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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