Current divider

Figure 4.1 shows an ideal current source injecting current into a node connected to two resistors. Both resistors connect from the same upper node to ground, so they are in parallel. Charge cannot disappear at the ideal node.

Kirchhoff's current law (KCL) states, for the chosen references,

This equation alone tells us that current divides, but not yet how much enters each branch. The missing information is the common branch voltage plus each branch's constitutive law.

Let V denote the upper-node voltage relative to ground. Because both resistors terminate on the same node pair, both branches have that same V. Ohm's law gives

where conductance is defined by G=1/R. At a fixed common voltage, a branch with twice the conductance carries twice the current. This is the most direct physical interpretation of current division.

The inverse resistance ratio appears because conductance is the reciprocal of resistance.

Insert the branch laws into KCL:

Therefore the common voltage is

Substitute this voltage back into each branch current:

This is the current-divider relation in its most transparent form: each branch receives the same fraction of total current as its conductance contributes to the total conductance.

For exactly two resistors, replace each conductance by its reciprocal resistance:

Multiply numerator and denominator by R1R2:

Similarly,

The “opposite” resistance appears only because reciprocals were cleared algebraically. The physical rule remains simpler: branch current is proportional to that branch's own conductance.

An ideal current source injects I=6 mA into R1=3 kΩ and R2=6 kΩ in parallel.

Step 1 — use the conductance fractions.

Step 2 — KCL check.

Step 3 — independent common-voltage check.

Both branches return exactly the same node voltage, as parallel topology requires.

Step 4 — power check.

The resistors absorb 72 mW in total, equal to the power delivered by the ideal source.

For N parallel branches with conductances G1 through GN, the current in branch k is

If all N resistors are equal, all conductances are equal and symmetry gives

Useful limiting checks for branch 1 are:

• R1 → 0: branch 1 approaches an ideal short, its conductance dominates, and I1 → I.
• R1 → ∞: branch 1 approaches an open circuit, its conductance vanishes, and I1 → 0.
• R1=R2: the two currents are equal, each carrying I/2.

These checks are often faster than redoing the algebra when debugging an answer.

Suppose branch 1 must carry a fraction β of the total current:

The conductance form gives

or, in resistance form,

Example: 75% / 25% split. If branch 1 must carry 75% of the total, β=0.75 and R2/R1=3. Choosing R1=2 kΩ and R2=6 kΩ produces the desired ratio. With a 4 mA ideal source, branch currents are 3 mA and 1 mA.

As with a voltage divider, the ratio determines the split but not the absolute resistance scale. That scale matters for source voltage and power.

Multiply every branch resistance by the same positive factor λ. The current fractions do not change because all conductances scale by the same reciprocal factor. But the parallel equivalent resistance scales by λ:

For an ideal current source, the node voltage becomes

and the total resistor power becomes

An ideal current source can supply whatever voltage is required. A real current source cannot: it has a finite compliance-voltage range, meaning the range of output voltage over which it can maintain the specified current.

A robust workflow is: identify the common nodes, write KCL, express branch currents with their element laws, solve for the common node voltage if necessary, then verify the current sum and physical limits.

1. A 9 mA source feeds 2 kΩ and 4 kΩ in parallel. Find both branch currents and the common voltage.
2. A 12 mA source must split equally among three identical branches. Find each branch current.
3. A two-branch divider must send 80% of the total current through branch 1. Find the required resistance ratio R2/R1.
4. A 5 mA source feeds R1=1 kΩ and R2=9 kΩ. Estimate the branch currents before calculating them exactly.
5. Two current-divider designs use the same resistance ratio: 1 kΩ/3 kΩ and 100 kΩ/300 kΩ. With the same ideal source current, compare the current fractions and required node voltages.
6. A real current source can maintain 2 mA only while its output voltage remains below 10 V. It feeds two equal 20 kΩ resistors in parallel. Will the ideal current-divider result be physically reachable?

The final chapter of DC-03 will combine these ideas into loading and equivalent-resistance reasoning: when a connected circuit changes the network you thought you had.