Figure 7.

Series-parallel circuit.

(1) The first step is to reduce the two parallel resistors, B and C, to an

equivalent single resistance.

As B and C are equal, divide 10 by 2,

which gives 5 ohms as the total parallel resistance.

The circuit is

now a simple series circuit of two 5-ohm resistors.

The total

resistance is obtained by adding the resistance of A to the equivalent

resistance of B and C.

This gives 5 plus 5, or 10 ohms, as the

resistance for the entire circuit.

(2) The total current is calculated by means of Ohm's law:

Total current I = E/R total = 10/10 = 1 ampere.

(3) This 1 ampere flows through resistor A, giving a voltage drop of 5

volts.

As the two parallel resistances have the same value, the 1

ampere of current divides equally between the two.

The voltage drop

across B equals I x R = 1/2 ampere times 10 ohms, or 5 volts. Since

resistors B and C are parallel, this is also the voltage drop across

resistor C.

(4) Following one complete path around the circuit, you can see that the

sum of the voltage drops is equal to the applied voltage.

Starting

from the positive side of the battery there is a 5-volt drop across

resistor A, another 5-volt drop across resistor B, and back to the

battery.

Following a path through resistor C will yield the same

results.

Care must be taken to follow only one path at a time when

tracing through a circuit.

b. Sample Calculation.

Refer to figure 8.

(1) Resistors V and W in series yield a total resistance of 10 ohms.

(2) Resistor B in parallel with series resistors V and W yields a parallel

resistance of 5 ohms.

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