A parallel circuit can now be established by connecting a supply across
a set of parallel resistors as shown in
Fig. 1. The positive terminal of
the supply is directly connected to the top of each resistor, while the
negative terminal is connected to the bottom of each resistor. Therefore,
it should be quite clear that the applied voltage is the same across each
resistor.
Fig. 1: Parallel Network.
In general,
the voltage is always the same across parallel elements.
Therefore, remember that
if two elements are in parallel, the voltage across them must be the
same. However, if the voltage across two neighboring elements is the
same, the two elements may or may not be in parallel.
For the voltages of the circuit in
Fig. 1, the result is that
Once the supply has been connected, a source current is established
through the supply that passes through the parallel resistors.
The smaller the total resistance, the greater is the current, as
occurred for series circuits also. The source current can then be determined
using Ohm's law:
Since the voltage is the same across parallel elements, the current
through each resistor can also be determined using
Ohm's law. That is,
$$ I_1 ={V_1 \over R_1} = {E \over R_1}$$
$$ I_2 = {V_2 \over R_2} = {E \over R_2}$$
The direction for the currents is dictated by the polarity of the voltage
across the resistors. Recall that for a resistor, current enters the positive
side of a potential drop and leaves the negative side. The result, as shown
in
Fig. 1, is that the source current enters point a, and currents $I_1$ and
$I_2$ leave the same point.
For single-source parallel networks, the source current (Is) is always
equal to the sum of the individual branch currents.
$$\bbox[5px,border:1px solid grey] {I_s = I_1 + I_2} $$
$$ {E \over R_T} = {V_1 \over R_1} + { V_2 \over R_2}$$
But $ E = V_1 = V_2$
$$ {E \over R_T} = {E \over R_1} + { E \over R_2}$$
$$ \bbox[5px,border:1px solid grey] {{1 \over R_T} = {1 \over R_1} + { 1 \over R_2}}$$
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