Network Theorems and Transformations: Current divider rule

Network Theorems and Transformations: Current divider rule

1. What Current Divider Is

Current divider = parallel circuit rule to find how current splits among parallel branches from total current entering node, without solving whole circuit.

For two resistors $R_1,R_2$ in parallel, total current $I_{total}$ entering node splits as:

i.e., current in a branch = total × opposite resistance / sum. Larger $R$ gets smaller share (inverse with $R$).

Derivation: $V = I_{total}·R_{eq}$ where $R_{eq}=R_1R_2/(R_1+R_2)$, then $I_1=V/R_1= I_{total}·R_2/(R_1+R_2)$.

For conductances $G=1/R$:

— proportional to $G$.

2. For More Than Two Branches

For $n$ parallel resistors $R_i$, current in $R_k$:

or

where

3. In Terms of Conductance (often easier)

For parallel, $G_{eq}= Σ G_i$,

4. Worked Examples

(a) I_total=10A into parallel 2Ω and 3Ω: I_2Ω=10·3/(2+3)=6A, I_3Ω=10·2/5=4A ✓ Check: 6+4=10, V=6·2=12V, 4·3=12V same.

(b) Three parallel 2Ω,3Ω,6Ω: G: 0.5,0.333,0.167 sum=1 S, R_eq=1Ω, V=10A·1=10V ⇒ I2=5A, I3≈3.33A, I6≈1.67A ✓

(c) Why not for series? Series uses voltage divider, not current.

(d) Need to find branch current quickly without nodal? Use divider.

(e) For inductors/capacitors? For AC, replace R with Z: I1= I_total·Z2/(Z1+Z2) (for parallel).

5. Exam Points

• State $I_1=I·R_2/(R_1+R_2)$ for two parallel — 5 marks, very frequent (almost certain to ask as divider).
• Derive via $V=I·R_{eq}$ — 3-5 marks, frequent.
• Solve numerical with 2 or 3 parallel — 5-7 marks, very frequent.
• Distinguish from voltage divider (series) — 3 marks.