Intuition The big pictureA current divider is what happens when a single current arrives at a junction and has two or more parallel paths to flow through. The current splits, but not equally — it prefers the path of least resistance .
WHY it matters: real circuits constantly branch. Knowing how current splits lets you size resistors, protect LEDs, and read meters without solving the whole network.
Definition Current dividerA network where a total current I T I_T I T parallel branches. Because parallel branches share the same voltage , the branch currents are set purely by each branch's resistance (or conductance).
Two facts do all the work:
KCL (Kirchhoff's Current Law): current in = current out, so I T = I 1 + I 2 I_T = I_1 + I_2 I T = I 1 + I 2 Parallel branches share one voltage V V V
Two resistors R 1 R_1 R 1 R 2 R_2 R 2 I T I_T I T
Step 1 — Same voltage.
Why? Both resistors connect between the same two nodes , so they see the identical voltage V V V
V = I 1 R 1 = I 2 R 2 V = I_1 R_1 = I_2 R_2 V = I 1 R 1 = I 2 R 2
Step 2 — Total current from Ohm's law on the equivalent resistance.
Why? The whole bundle behaves like one resistor R ∥ R_\parallel R ∥ I T I_T I T V V V
V = I T R ∥ , R ∥ = R 1 R 2 R 1 + R 2 V = I_T\,R_\parallel,\qquad R_\parallel=\frac{R_1 R_2}{R_1+R_2} V = I T R ∥ , R ∥ = R 1 + R 2 R 1 R 2
Step 3 — Find I 1 I_1 I 1
Why? I 1 = V / R 1 I_1 = V/R_1 I 1 = V / R 1 V = I T R ∥ V=I_T R_\parallel V = I T R ∥
I 1 = V R 1 = I T R ∥ R 1 = I T R 1 ⋅ R 1 R 2 R 1 + R 2 I_1=\frac{V}{R_1}=\frac{I_T R_\parallel}{R_1}=\frac{I_T}{R_1}\cdot\frac{R_1R_2}{R_1+R_2} I 1 = R 1 V = R 1 I T R ∥ = R 1 I T ⋅ R 1 + R 2 R 1 R 2
The R 1 R_1 R 1
Sanity check: add them.
I 1 + I 2 = I T R 2 + R 1 R 1 + R 2 = I T ✓ I_1+I_2 = I_T\frac{R_2+R_1}{R_1+R_2}=I_T \checkmark I 1 + I 2 = I T R 1 + R 2 R 2 + R 1 = I T ✓
Intuition Why conductance is cleanerResistance says "how hard to push"; conductance G = 1 / R G=1/R G = 1/ R loves easy paths, so it splits in direct proportion to conductance — no confusing swaps.
Since I k = V G k I_k = V G_k I k = V G k I T = V ∑ G j I_T = V\sum G_j I T = V ∑ G j
For exactly two resistors this collapses back to the boxed formula (try it — see Verify).
Worked example 1 — Simple two-resistor splitI T = 12 mA I_T = 12\text{ mA} I T = 12 mA R 1 = 2 k Ω R_1 = 2\text{ k}\Omega R 1 = 2 k Ω R 2 = 4 k Ω R_2 = 4\text{ k}\Omega R 2 = 4 k Ω I 1 = 12 ⋅ R 2 R 1 + R 2 = 12 ⋅ 4 6 = 8 mA I_1 = 12\cdot\frac{R_2}{R_1+R_2}=12\cdot\frac{4}{6}=8\text{ mA} I 1 = 12 ⋅ R 1 + R 2 R 2 = 12 ⋅ 6 4 = 8 mA Why R 2 R_2 R 2 I 1 I_1 I 1 R 1 R_1 R 1 I 2 = 12 ⋅ 2 6 = 4 mA . 8 + 4 = 12 ✓ I_2 = 12\cdot\frac{2}{6}=4\text{ mA}.\quad 8+4=12\checkmark I 2 = 12 ⋅ 6 2 = 4 mA . 8 + 4 = 12 ✓
Worked example 2 — Forecast then verifyI T = 9 mA I_T = 9\text{ mA} I T = 9 mA R 1 = 1 k Ω R_1 = 1\text{ k}\Omega R 1 = 1 k Ω R 2 = 8 k Ω R_2 = 8\text{ k}\Omega R 2 = 8 k Ω Forecast: R 1 R_1 R 1 I 1 = 9 ⋅ 8 9 = 8 mA , I 2 = 9 ⋅ 1 9 = 1 mA I_1 = 9\cdot\frac{8}{9}=8\text{ mA},\qquad I_2=9\cdot\frac{1}{9}=1\text{ mA} I 1 = 9 ⋅ 9 8 = 8 mA , I 2 = 9 ⋅ 9 1 = 1 mA Why this step? The ratio R 2 R 1 + R 2 = 8 9 \frac{R_2}{R_1+R_2}=\frac{8}{9} R 1 + R 2 R 2 = 9 8
Worked example 3 — Three branches (conductance form)I T = 30 mA I_T = 30\text{ mA} I T = 30 mA R 1 = 10 Ω R_1=10\,\Omega R 1 = 10 Ω R 2 = 20 Ω R_2=20\,\Omega R 2 = 20 Ω R 3 = 20 Ω R_3=20\,\Omega R 3 = 20 Ω G 1 = 0.1 , G 2 = 0.05 , G 3 = 0.05 G_1=0.1,\ G_2=0.05,\ G_3=0.05 G 1 = 0.1 , G 2 = 0.05 , G 3 = 0.05 ∑ G = 0.2 \sum G = 0.2 ∑ G = 0.2 I 1 = 30 ⋅ 0.1 0.2 = 15 mA I_1 = 30\cdot\frac{0.1}{0.2}=15\text{ mA} I 1 = 30 ⋅ 0.2 0.1 = 15 mA Why? R 1 R_1 R 1 I 2 = I 3 = 30 ⋅ 0.05 0.2 = 7.5 mA . 15 + 7.5 + 7.5 = 30 ✓ I_2=I_3=30\cdot\frac{0.05}{0.2}=7.5\text{ mA}.\quad 15+7.5+7.5=30\checkmark I 2 = I 3 = 30 ⋅ 0.2 0.05 = 7.5 mA . 15 + 7.5 + 7.5 = 30 ✓
Common mistake "Current is bigger through the bigger resistor — putR 1 R_1 R 1 I 1 I_1 I 1 Why it feels right: in a voltage divider you take the resistor you care about over the total, so people copy that pattern. It also feels like "bigger resistor, bigger share."
The fix: current is the opposite — it avoids big resistors. For I 1 I_1 I 1 other resistor R 2 R_2 R 2 I 1 = V / R 1 I_1=V/R_1 I 1 = V / R 1 R 1 R_1 R 1 smaller I 1 I_1 I 1
Common mistake "For 3+ resistors I'll just useI 1 = I T R 2 / ( R 1 + R 2 ) I_1 = I_T R_2/(R_1+R_2) I 1 = I T R 2 / ( R 1 + R 2 ) Why it feels right: the two-resistor formula is memorable and everywhere.
The fix: the "swap the other resistor" trick only works for two branches. For N ≥ 3 N\ge3 N ≥ 3 I k = I T G k / ∑ G j I_k = I_T\,G_k/\sum G_j I k = I T G k / ∑ G j
Common mistake "Resistors in series form a current divider."Why it feels right: series and parallel look similar on a messy schematic.
The fix: series shares current (same current everywhere) and divides voltage . Only parallel branches divide current.
Recall Feynman: explain to a 12-year-oldImagine water reaching a fork with two pipes. If one pipe is wide (low resistance) and one is skinny (high resistance), more water rushes through the wide pipe. The total water stays the same, it just splits — and the wide, easy pipe grabs the bigger share. Current in a wire does exactly this: it flows most through the easiest path.
Mnemonic Remember the swap"CROSS to find your current." For I 1 I_1 I 1 R 2 R_2 R 2 "Current takes the easy road" — low-R R R
Why do parallel branches share the same voltage?
In I 1 = I T R 2 / ( R 1 + R 2 ) I_1 = I_T R_2/(R_1+R_2) I 1 = I T R 2 / ( R 1 + R 2 ) R 2 R_2 R 2 R 1 R_1 R 1
What form must you use for 3+ branches, and why?
Current divider requires branches to be in — series or parallel? Parallel (they share the same voltage).
Two-resistor current divider formula for I 1 I_1 I 1 I 1 = I T R 2 / ( R 1 + R 2 ) I_1 = I_T\,R_2/(R_1+R_2) I 1 = I T R 2 / ( R 1 + R 2 ) — the
other resistor goes on top.
Why is the "other" resistor on top for a current divider? Because
I 1 = V / R 1 = I T R ∥ / R 1 I_1=V/R_1=I_T R_\parallel/R_1 I 1 = V / R 1 = I T R ∥ / R 1 , and
R ∥ R_\parallel R ∥ contains
R 2 R_2 R 2 ; the
R 1 R_1 R 1 cancels leaving
R 2 R_2 R 2 on top. Current avoids large resistance.
General N-branch current divider formula I k = I T G k / ∑ j G j I_k = I_T\,G_k/\sum_j G_j I k = I T G k / ∑ j G j with
G = 1 / R G=1/R G = 1/ R ; current splits in proportion to conductance.
Does current or voltage divide in a parallel network? Current divides; voltage is common to all branches.
Which branch carries the most current in a divider? The branch with the lowest resistance (highest conductance).
Check that branch currents sum correctly I 1 + I 2 = I T ( R 2 + R 1 ) / ( R 1 + R 2 ) = I T I_1+I_2 = I_T(R_2+R_1)/(R_1+R_2)=I_T I 1 + I 2 = I T ( R 2 + R 1 ) / ( R 1 + R 2 ) = I T — satisfies KCL.
I_k = I_T times G_k / sum G_j
Junction with parallel branches
R_parallel = R1 R2 / (R1+R2)
I_1 = I_T times R2/(R1+R2)
Other resistor sits on top
N-branch conductance form
Worked examples verify sum = I_T
Intuition Hinglish mein samjhoDekho, current divider ka funda simple hai: jab total current I T I_T I T parallel raaste milte hain, toh current baant jaata hai — lekin barabar nahi. Current hamesha easy path (kam resistance) ko prefer karta hai, jaise paani mota pipe pehle bharta hai.
Do resistor wale case mein magic formula hai: I 1 = I T ⋅ R 2 / ( R 1 + R 2 ) I_1 = I_T \cdot R_2/(R_1+R_2) I 1 = I T ⋅ R 2 / ( R 1 + R 2 ) I 1 I_1 I 1 doosra resistor R 2 R_2 R 2 R 1 R_1 R 1 V V V I 1 = V / R 1 I_1 = V/R_1 I 1 = V / R 1 V = I T R ∥ V = I_T R_\parallel V = I T R ∥ R 1 R_1 R 1 R 2 R_2 R 2 do branch ke liye chalta hai.
Teen ya zyada branch ho toh formula ko conductance (G = 1 / R G = 1/R G = 1/ R I k = I T ⋅ G k / ∑ G j I_k = I_T \cdot G_k/\sum G_j I k = I T ⋅ G k / ∑ G j
Sabse common galti: log R 1 R_1 R 1 current bade resistor se bhaagta hai , isliye chhote resistor wali branch mein zyada current. Exam mein hamesha check karo: I 1 + I 2 = I T I_1 + I_2 = I_T I 1 + I 2 = I T
