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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