Build and analyze a current divider: Question bank

For the machinery being tested here, keep these prerequisites in view: Kirchhoff's Current Law, Ohm's Law, Parallel resistance, Conductance, and the series-dual Voltage divider.

True or false — justify

True or false: In a current divider the branch with the larger resistance carries the larger current.

False. Current is

I = V / R

with

V

shared, so a larger

R

means a smaller current — the easy (low-

R

) path grabs more.

True or false: The two resistors in a current divider must share the same voltage.

True. They connect between the same two nodes, and voltage is measured between nodes — so both see the identical

V

. That shared

V

is what makes the split depend only on resistance.

True or false: Branch currents in a divider always add up to the total current

I_{T}

True. This is just Kirchhoff's Current Law — charge is conserved at the node, so

I_{1} + I_{2} + \dots = I_{T}

no matter the resistances.

True or false: For

I_{1}

you put

R_{1}

on top:

I_{1} = I_{T} R_{1} / (R_{1} + R_{2})

False. It's the other resistor:

I_{1} = I_{T} R_{2} / (R_{1} + R_{2})

. Confusing it with the Voltage divider (where you do use your own resistor) is the classic trap.

True or false: Current splits in direct proportion to each branch's conductance

G = 1/ R

True. Since

I_{k} = V G_{k}

and

V

is shared,

I_{k} \propto G_{k}

— the easier the path (bigger

G

), the more current. This is the cleanest way to see the whole thing.

True or false: Two equal resistors in parallel split the current exactly in half.

True. Equal

R

means equal conductance, so each takes the same share

\frac{1}{2} I_{T}

— the only case where the divider is "fair."

True or false: Making

R_{2}

larger increases the current through

R_{1}

True (with

I_{T}

held fixed).

I_{1} = I_{T} R_{2} / (R_{1} + R_{2})

grows toward

I_{T}

R_{2} \to \infty

— a fatter "other" resistor pushes more current to stay in your branch.

True or false: Resistors in series form a current divider.

False. Series branches carry the same current and divide voltage; only parallel branches divide current.

True or false: The two-resistor formula

I_{1} = I_{T} R_{2} / (R_{1} + R_{2})

still works for three parallel branches if you just ignore the third.

False. Ignoring the third branch changes where the current goes. For

N \geq 3

you must use

I_{k} = I_{T} G_{k} / \sum_{j} G_{j}

Spot the error

I_{1} = I_{T} \cdot \frac{R _{1}}{R _{1} + R _{2}}

because

I_{1}

flows through

R_{1}

, so

R_{1}

belongs on top."

Error: the resistor you flow through goes on the bottom of

I = V / R

. Derive it —

I_{1} = V / R_{1} = I_{T} R_{∥} / R_{1}

, and

R_{∥} = R_{1} R_{2} / (R_{1} + R_{2})

, so the

R_{1}

cancels leaving

R_{2}

on top.

"With three branches,

I_{1} = I_{T} \frac{R _{2} R _{3}}{R _{1} + R _{2} + R _{3}}

— just extend the pattern."

Error: neither the numerator nor the denominator is right. Use conductances:

I_{1} = I_{T} G_{1} / (G_{1} + G_{2} + G_{3})

with

G = 1/ R

. The "swap" trick has no clean generalization.

"Since

V = I_{1} R_{1}

and

V = I_{2} R_{2}

, and both equal

V

, we get

I_{1} = I_{2}

Error:

I_{1} R_{1} = I_{2} R_{2}

does not give

I_{1} = I_{2}

unless

R_{1} = R_{2}

. It gives

I_{1} / I_{2} = R_{2} / R_{1}

— the currents are inversely proportional to the resistances.

"The equivalent resistance

R_{∥} = R_{1} + R_{2}

, so I'll use that for

V = I_{T} R_{∥}

Error: that's the series formula. In parallel

R_{∥} = R_{1} R_{2} / (R_{1} + R_{2})

, which is always smaller than either resistor — adding a path makes flow easier, not harder.

I_{2} = I_{T} - I_{1}

, but also

I_{2} = I_{T} R_{1} / (R_{1} + R_{2})

, and these disagree."

No error — they must agree.

I_{T} - I_{T} R_{2} / (R_{1} + R_{2}) = I_{T} (R_{1} + R_{2} - R_{2}) / (R_{1} + R_{2}) = I_{T} R_{1} / (R_{1} + R_{2})

. KCL and the divider formula are consistent by construction.

"A perfect wire (0 Ω) placed in parallel with

R_{1}

still lets some current through

R_{1}

Error: a

0 Ω

branch has infinite conductance, so its share

G / \sum G \to 1

— it grabs essentially all the current, shorting

R_{1}

out. Practically no current flows through

R_{1}

Why questions

Why does current "prefer" the low-resistance branch instead of splitting equally?

Because branches share one voltage

V

, and

I = V / R

: the smaller the

R

, the bigger the

I

for that same push. Nothing "decides" — it's forced by Ohm's Law.

Why is conductance

G = 1/ R

the more natural variable for current division?

Because

I_{k} = V G_{k}

makes current directly proportional to

G

, so the split

I_{k} / I_{T} = G_{k} / \sum G

has no confusing swap — the biggest

G

simply takes the biggest slice.

Why does the

R_{1}

cancel in the derivation, leaving only

R_{2}

on top?

We compute

I_{1} = V / R_{1}

with

V = I_{T} R_{∥}

and

R_{∥} = R_{1} R_{2} / (R_{1} + R_{2})

. The

R_{1}

R_{∥}

cancels the

R_{1}

in the denominator, leaving

I_{1} = I_{T} R_{2} / (R_{1} + R_{2})

Why does the current divider look "backwards" compared to the Voltage divider?

In a voltage divider (series) you keep your own resistor on top; in a current divider (parallel) you use the other one. The duality: voltage divides across series resistances, current divides across parallel conductances.

Why does adding a parallel branch decrease the equivalent resistance?

Each new path is another way for current to flow, so total conductance

\sum G

only grows — and

R_{∥} = 1/ \sum G

shrinks. More doors always make a room easier to leave.

Why can two resistors of very different sizes still be called a "divider" even if one branch gets almost nothing?

Because the current still splits — it's just a lopsided split. A

1 k Ω

1 M Ω

pair sends ~99.9% down the small one, but both branches carry some current.

Edge cases

Edge case: what happens to

I_{1}

R_{2} \to \infty

(its branch becomes an open circuit)?

I_{1} = I_{T} R_{2} / (R_{1} + R_{2}) \to I_{T}

. With the other branch open, all the current is forced through

R_{1}

— the divider degenerates into a single path.

Edge case: what happens to

I_{1}

R_{2} \to 0

(its branch becomes a short)?

I_{1} = I_{T} R_{2} / (R_{1} + R_{2}) \to 0

. The

0 Ω

branch has infinite conductance and hogs everything, so

I_{1}

vanishes —

R_{1}

is bypassed.

Edge case: what does the formula give if

R_{1} = R_{2}

I_{1} = I_{T} R / (2 R) = I_{T} /2

, and likewise

I_{2} = I_{T} /2

. Equal resistances, equal split — the sanity anchor for the whole formula.

Edge case: what if

I_{T} = 0

(no current enters the node)?

Then

I_{1} = I_{2} = 0

: the divider ratio

R_{2} / (R_{1} + R_{2})

is still well-defined, but there's simply nothing to divide. The proportions hold, the magnitudes are zero.

Edge case: is there ever a resistance choice that makes a branch current exceed

I_{T}

No. Each ratio

R_{2} / (R_{1} + R_{2})

and

G_{k} / \sum G

lies strictly between

0

and

1

, so every branch current is a fraction of

I_{T}

. Current can only be shared, never amplified, by passive resistors.

Edge case: what if two branches have equal resistance but a third is much smaller — how does the small one affect the equal pair?

The small branch grabs the lion's share (largest

G

), leaving less for the equal pair — but the two equal branches still split their remainder equally, since

G_{2} = G_{3}

keeps

I_{2} = I_{3}

Recall One-line summary to lock in

Current divides in a parallel network in proportion to conductance ( $I_{k} = I_{T} G_{k} / \sum G$ ); for two branches the shortcut is $I_{1} = I_{T} R_{2} / (R_{1} + R_{2})$ with the other resistor on top — and every branch current is a fraction of $I_{T}$ , summing back to $I_{T}$ by Kirchhoff's Current Law.

Connections

Kirchhoff's Current Law — guarantees the branch currents sum to $I_{T}$ .
Voltage divider — the series dual whose "own resistor on top" rule people wrongly copy.
Parallel resistance — $R_{∥}$ is the object that makes $R_{1}$ cancel in the derivation.
Ohm's Law — $I = V / R$ per branch is the whole reason big $R$ means small $I$ .
Conductance — $G = 1/ R$ , the variable that makes the split proportional and clean.