Compute equivalent resistance in mixed networks: Question bank

Before we start, one word we'll lean on: a node is just a single electrical junction — a point (or an unbroken piece of wire) where component ends meet. Two points joined by a plain wire with nothing in between are the same node. Keep that picture close; half these traps are really "how many nodes touch here?" in disguise.

True or false — justify

Two resistors that are physically drawn side by side on the schematic are in parallel

False — layout on paper means nothing; parallel requires sharing both end nodes. They could share zero nodes and be completely unrelated.

Series resistance is always larger than either individual resistor

True — adding a resistor lengthens the single current path, so

R_{ser i es} = R_{1} + R_{2}

exceeds each part; there is no way for a sum of positive resistances to be smaller than a term.

Parallel resistance is always smaller than the smallest branch

True — extra branches only give current more roads, never fewer, so the effective resistance drops below even the tightest single road.

If you connect two identical resistors in parallel, the equivalent is half of one

True —

\frac{R \cdot R}{R + R} = \frac{R}{2}

; two equal roads split the current evenly, doubling total flow for the same voltage, which halves resistance.

Adding more resistors in parallel can eventually make the equivalent resistance negative

False — each added conductance

1/ R

is positive, so

R_{e q}

keeps shrinking toward zero but stays strictly positive; resistance of ordinary resistors is never negative.

A wire (zero-ohm connection) placed in parallel with a resistor leaves that resistor's current unchanged

False — the

0 Ω

path is a short;

R_{e q} = \frac{R \cdot 0}{R + 0} = 0

, so essentially all current diverts through the wire and bypasses the resistor.

The order in which you collapse series/parallel groups changes the final

R_{e q}

False — because the box behaves as one fixed

R_{e q}

(Ohm's law is linear, see Ohm's Law), any valid collapse order lands on the same number; only your bookkeeping effort changes.

Every two-terminal resistor network can be reduced by repeatedly applying only the series and parallel rules

False — bridge networks have no purely series or parallel pair to grab; they need a Delta-Wye Transformation first.

If two resistors carry the same current, they must be in series

False — series guarantees equal current, but equal current does not guarantee series. Two matched resistors in symmetric parallel branches (see the right panel of the figure) carry identical current yet are not in series; you must still check the shared node touches nothing else.

Doubling every resistor in a network doubles

R_{e q}

True — both the series sums and the parallel reciprocal sums scale linearly, so multiplying all resistances by a constant multiplies

R_{e q}

by that same constant.

Spot the error

R_{1}

and

R_{2}

meet at node C, and a third resistor

R_{3}

also connects to C, so

R_{1}

and

R_{2}

are in series." — find the flaw

The extra branch at C splits the current (KCL, see Kirchhoff's Current Law), so

R_{1}

and

R_{2}

no longer carry the same current — series requires the shared node to touch nothing else.

"Two parallel resistors:

\frac{1}{R _{1}} + \frac{1}{R _{2}} = 0.5

, so

R_{e q} = 0.5 Ω

." — find the flaw

That

0.5

is the conductance

1/ R_{e q}

, not the resistance; you must flip it, giving

R_{e q} = 1/0.5 = 2 Ω

R_{1} = 3 Ω ∥ R_{2} = 6 Ω

gives

R_{e q} = 9 Ω

." — find the flaw

They added resistances (a series move) for a parallel pair; a parallel result must be below the smallest branch (

< 3 Ω

), so

9 Ω

trips the "parallel must shrink" bug alarm — the real answer is

2 Ω

"For three resistors in parallel I used product-over-sum:

\frac{R _{1} R _{2} R _{3}}{R _{1} + R _{2} + R _{3}}

." — find the flaw

Product-over-sum is derived only for two resistors; for three you must return to the reciprocal sum

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}}

"A branch has no current flowing, so I can ignore that resistor and delete it." — find the flaw

You may delete a branch only if it's a genuine open (broken); a branch with zero current in the present solution may still carry current if the rest of the circuit changes, and deleting it changes the topology you're analyzing.

"These two resistors share both nodes, so they're in series." — find the flaw

Sharing both nodes is the parallel condition (same voltage); series is sharing exactly one node with nothing else attached. The student swapped the two definitions.

"I merged

R_{1}

and

R_{2}

as parallel, then re-used the original

R_{1}

separately later." — find the flaw

Once you collapse a group into an equivalent, the originals are gone from the redrawn circuit; reusing

R_{1}

double-counts it and corrupts every later step.

Why questions

Why does KVL justify the series-addition rule?

Around the single path the voltage drops must add up to the total applied voltage (see Kirchhoff's Voltage Law); with the same current

I

V = I R_{1} + I R_{2} = I (R_{1} + R_{2})

, so the equivalent resistance is the sum.

Why does KCL justify the parallel-addition rule?

At the shared node the branch currents add to the total (see Kirchhoff's Current Law); with the same voltage

V

I = \frac{V}{R _{1}} + \frac{V}{R _{2}}

, so the conductances add and

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

Why is it natural to add conductances rather than resistances in parallel?

Conductance

G = 1/ R

measures how easily current flows; parallel branches offer independent extra roads, so their ease-of-flow simply piles up — the additive quantity is

G

, as Conductance and Admittance makes explicit.

Why does the linearity of Ohm's law let us swap any sub-network for a single resistor?

Linearity means the box's terminal behaviour is exactly

V = I R_{e q}

with one fixed constant; nothing outside the box can tell the difference, so the substitution is always legal (Ohm's Law).

Why do we usually collapse the group farthest from the terminals first?

The innermost pair is the only place where a clean series or parallel relationship is fully "self-contained" (nothing else taps it yet); collapsing it exposes the next clean pair, working outward like peeling an onion.

Why is a bridge network immune to plain series/parallel reduction?

Its middle "bridge" resistor connects two nodes that each carry other branches, so no pair shares a node-with-nothing-else (series) nor shares both nodes (parallel) — you must reshape it with a Delta-Wye Transformation.

Why can the voltage-divider and current-divider rules be seen as shortcuts of these two atomic rules?

A Voltage Divider is just series resistors sharing one current, and a Current Divider is just parallel resistors sharing one voltage — the divider ratios fall straight out of

V = I R

applied to those configurations.

Edge cases

What is

R_{e q}

of a single resistor

R

in parallel with an ideal wire (

0 Ω

Zero — the wire short-circuits the resistor, so the pair conducts as if the resistor weren't there; all current takes the free road.

What is

R_{e q}

of a resistor

R

in series with an ideal wire (

0 Ω

Just

R

— a

0 Ω

series element adds nothing to the path length, so it's like a plain piece of connecting wire.

What is

R_{e q}

of a resistor

R

in parallel with an open (infinite resistance)?

Exactly

R

— an open branch carries no current, so it contributes zero conductance and leaves the resistor to do all the work.

What is

R_{e q}

of a resistor

R

in series with an open (infinite resistance)?

Infinite (open) — a break anywhere in a single path stops all current, so the whole series combination is an open circuit.

As one branch of a two-resistor parallel pair grows toward infinity, what does

R_{e q}

approach?

It approaches the other branch's resistance — the huge branch stops carrying meaningful current, so the finite branch alone sets

R_{e q}

As one branch of a two-resistor parallel pair shrinks toward zero, what does

R_{e q}

approach?

It approaches zero — the near-short branch hogs the current and pulls the equivalent resistance down with it, no matter how large the other branch is.

If two nodes labelled with different letters are actually joined by a bare wire, what should you do before analyzing?

Merge them into a single node (same electrical point) — otherwise you may falsely see two components as separated when they truly share a node.

What is the equivalent resistance of

n

identical resistors

R

all in parallel?

R / n

— the conductances add to

n / R

, and flipping gives

R_{e q} = R / n

; every added identical road cuts the resistance further.

Recall One-line self-audit after any collapse step

After each merge, ask two things: (1) did a parallel step leave me a value smaller than the smallest branch? and (2) did a series step leave me a value larger than every part? If either check fails, you mislabelled the group.

Connections

Parent: Compute equivalent resistance in mixed networks
1.2.02 Compute equivalent resistance in mixed networks (Hinglish)
Ohm's Law — linearity behind "one box = one resistor"
Kirchhoff's Voltage Law — series justification
Kirchhoff's Current Law — parallel justification
Delta-Wye Transformation — for bridge networks
Voltage Divider · Current Divider — divider shortcuts
Conductance and Admittance — why parallel adds $1/ R$