Encyclopedia of Electrical Engineering

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.
#### What is Current Divider Rule?

In general,
**Fig.no.1: **Discussing the manner in which the current
will split between three parallel branches
of different resistive value.
In the previous section, it was pointed out that current will always seek the
path of least resistance. In Fig. No.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.
**Example 1:** a. Determine the current I1 for the network of Fig. No.2 using the ratio
rule.

b. Determine the current I3 for the network of Fig. 2 using the ratio rule.

c. Determine the current Is using Kirchhoff's current law. **Fig.no.2: ** Parallel network for Example 1.
**Solution:**

a. Applying the ratio rule: $${I_1 \over I_2}={R_2 \over R_1}$$ $${I_1 \over 2mA}={3Ω \over 6Ω}$$ $$I_1 ={1Ω \over 2Ω} (2mA)$$ $$I_1 = 1mA $$ b. Applying the ratio rule: $${I_2 \over I_3}={R_3 \over R_2}$$ $${2mA \over I_3}={1Ω \over 3Ω}$$ $$I_3 =3 (2mA)$$ $$I_3 = 6mA $$ c. Applying Kirchhoff's current law: $$\sum{I_i} = \sum{I_o}$$ $$ I_s = I_1+I_2+I_3$$ $$ I_s = 1mA + 2mA + 6mA$$
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. No.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. No.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.
**Fig.No.3: **Deriving the current divider rule:

(a) parallel network of N parallel resistors;

(b) reduced equivalent of part (a).
The current $I_T$ can then be determined using Ohm's law:
$$I_T = {V \over R_T}$$
Since the voltage V is the same across parallel elements, the following
is true:
$$V = I_1 R_1 = I_2 R_2 = I_3 R_3 = . . . = I_x R_x$$
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
$$I_T = {I_x R_x \over RT}$$
Solving for $I_x$, the final result is the current divider rule:

which states that
#### Special Case: Two Parallel Resistors

For the case of two parallel resistors, the total
resistance is determined by
$$ R_T = {R_1 R_2 \over R_1 + R_2}$$
Substituting $R_T$ into Eq. (2) for current $I_1$ results in
$$ I_1 = {({R_1 R_2 \over R_1 + R_2)} \over R_1} I_T$$
and
$$ I_1 = ({R_2 \over R_1 + R_2}) I_T$$
Similarly, for I2,
$$ I_2 = ({R_1 \over R_1 + R_2}) I_T$$

$$\bbox[5px,border:1px solid red] {\color{blue}{{I_1 \over I_2}= {R_2 \over R_1}}}$$ | Eq.(1) |

b. Determine the current I3 for the network of Fig. 2 using the ratio rule.

c. Determine the current Is using Kirchhoff's current law.

a. Applying the ratio rule: $${I_1 \over I_2}={R_2 \over R_1}$$ $${I_1 \over 2mA}={3Ω \over 6Ω}$$ $$I_1 ={1Ω \over 2Ω} (2mA)$$ $$I_1 = 1mA $$ b. Applying the ratio rule: $${I_2 \over I_3}={R_3 \over R_2}$$ $${2mA \over I_3}={1Ω \over 3Ω}$$ $$I_3 =3 (2mA)$$ $$I_3 = 6mA $$ c. Applying Kirchhoff's current law: $$\sum{I_i} = \sum{I_o}$$ $$ I_s = I_1+I_2+I_3$$ $$ I_s = 1mA + 2mA + 6mA$$

(a) parallel network of N parallel resistors;

(b) reduced equivalent of part (a).

$$\bbox[5px,border:1px solid red] {\color{blue}{I_x = {I_T \over R_x} RT}}$$ | Eq.(2) |

Kirchhoffs Current law
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