Current Divider Rule
For series circuits we have the powerful voltage divider rule for finding
the voltage across a resistor in a series circuit. We now introduce the
equally powerful current divider rule (CDR) for finding the current
through a resistor in a parallel circuit.
In the previous section, it was pointed out that current will always seek the
path of least resistance. In Fig. 1, for example, the current of $9 A$ is
faced with splitting between the three parallel resistors. Based on the
previous sections, it should now be clear without a single calculation
that the majority of the current will pass through the smallest resistor of
$10 Ω$, and the least current will pass through the $1 kΩ$ resistor. In fact,
the current through the $100 Ω$ resistor will also exceed that through the
$1 kΩ$ resistor. We can take it one step further by recognizing that the
resistance of the $100 Ω$ resistor is 10 times that of the $10 Ω$ resistor.
The result is a current through the $10 Ω$ resistor that is 10 times that of
the 100 Ω resistor. Similarly, the current through the $100 Ω$ resistor is
10 times that through the $1 kΩ$ resistor.
Ratio Rule: Each of the panel statements above is supported by the
ratio rule, which states that for parallel resistors the current will divide as
the inverse of their resistor values.
In equation form:
The next example will demonstrate how quickly currents can be
determined using this important relationship.
The current $I_T$ can then be determined using Ohm's law:
Since the voltage V is the same across parallel elements, the following
is true:
where the product $I_x R_x$ refers to any combination in the series.
Substituting for $V$ in the above equation for $I_T$, we have
Solving for $I_x$, the final result is the current divider rule:
which states that
Substituting $R_T$ into Eq. (2) for current $I_1$ results in
and
Similarly, for $I_2$,
What is Current Divider Rule?
In general,
Fig. 1: Discussing the manner in which the current
will split between three parallel branches
of different resistive value.
$$\bbox[5px,border:1px solid grey] {{I_1 \over I_2}=
{R_2 \over R_1}} \tag{1}$$
Example 1:
a. Determine the current $I_1$ for the network of Fig. 2 using the ratio rule.
b. Determine the current $I_3$ for the network of Fig. 2 using the ratio rule.
c. Determine the current $Is$ using Kirchhoff's current law.
Solution:
a. Applying the ratio rule:
b. Applying the ratio rule:
c. Applying Kirchhoff's current law:
Although the above discussions and example allowed us to determine
the relative magnitude of a current based on a known level, they do
not provide the magnitude of a current through a branch of a parallel
network if only the total entering current is known. The result is a need
for the current divider rule, which will be derived using the parallel configuration
in Fig. 3(a). The current $I_T$ (using the subscript T to indicate
the total entering current) splits between the N parallel resistors and
then gathers itself together again at the bottom of the configuration. In
Fig. 3(b), the parallel combination of resistors has been replaced by a single resistor equal to the total resistance of the parallel combination as
determined in the previous sections.
a. Determine the current $I_1$ for the network of Fig. 2 using the ratio rule.
b. Determine the current $I_3$ for the network of Fig. 2 using the ratio rule.
c. Determine the current $Is$ using Kirchhoff's current law.
Fig. 2: Parallel network for Example 1.
a. Applying the ratio rule:
$${I_1 \over I_2}={R_2 \over R_1}$$
$${I_1 \over 2mA}={3Ω \over 6Ω}$$
$$I_1 ={1Ω \over 2Ω} (2mA)$$
$$I_1 = 1mA $$
$${I_2 \over I_3}={R_3 \over R_2}$$
$${2mA \over I_3}={1Ω \over 3Ω}$$
$$I_3 =3 (2mA)$$
$$I_3 = 6mA $$
$$\sum{I_i} = \sum{I_o}$$
$$ I_s = I_1+I_2+I_3$$
$$ I_s = 1mA + 2mA + 6mA$$
Fig. 3: Deriving the current divider rule:
(a) parallel network of N parallel resistors;
(b) reduced equivalent of part (a).
(a) parallel network of N parallel resistors;
(b) reduced equivalent of part (a).
$$I_T = {V \over R_T}$$
$$V = I_1 R_1 = I_2 R_2 = I_3 R_3 = . . . = I_x R_x$$
$$I_T = {I_x R_x \over RT}$$
$$\bbox[5px,border:1px solid grey] {I_x = {I_T \over R_x} R_T} \tag{2}$$
Special Case: Two Parallel Resistors
For the case of two parallel resistors, the total resistance is determined by$$ R_T = {R_1 R_2 \over R_1 + R_2}$$
$$ I_1 = {({R_1 R_2 \over R_1 + R_2)} \over R_1} I_T$$
$$ I_1 = ({R_2 \over R_1 + R_2}) I_T$$
$$ I_2 = ({R_1 \over R_1 + R_2}) I_T$$
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