Tricky Current Divider
Spring 2019
Current Divider Rule
Use CDR with a tricky circuit
Learn It!
Pre-Requisite Knowledge
Using CDR
Using the Current Divider Rule and other tools you have, find the value of I in the circuit below
\[ \begin{equation} a. 18 mA \\b. 11 mA \\c. 9mA \\d. 6mA \\e. 4mA \end{equation} \]
Part 1
Using CDR
Choose a solution strategy
\(I_1=I_{total} \dfrac{R_1}{R_1+R_2}\)
This is the Current Divider Rule (CDR).
\(R_2\) is the equivalent resistance of the rest of the resistors parallel to \(R_1\)
Currents in parallel resistors have this special relationship.
The way the CDR is usually written (applied to the leftmost resistor in a parallel) , it would output the current through the \(2 \Omega\) resistor. but what we want is the total current going through the \(3 \Omega\) and \(6 \Omega\) resistors.

That's OK, though, we can use that information to get what we want using KCL.
\(R_{eq}=\dfrac{1}{ \dfrac{1}{3}+\dfrac{1}{6} }=2\)
We need the equivalent resistance of the right two resistors to make this happen.
By coincidence, the equivalent resistance of the two rightmost resistors is exactly that of the left resistor.

In this situation, we won't have to use KCL after all.
\[I= 11mA\]
When two equal valued resistors are in parallel, they split the current evenly. Since we are splitting \(22mA\) in half, both halves of the circuit get \(11mA\).
It is important to take advantage of special features of particular problems, it can save precious minutes on an exam. If the left resistor were \(2.1\) instead of \(2\), we would not have been able to use this time-saving trick.