Series and parallel resistance: Question bank — Physics deep dive

Definition Three symbols this page leans on

Conductance $G$ is simply the reciprocal of resistance: $G = \frac{1}{R}$ , measured in siemens. It means "how easily current flows" — the exact opposite of resistance. Where resistors fight current, conductance measures the ease. Parallel branches add their $G$ because opening more roads adds more ease.
The summation sign $\sum$ is shorthand for "add them all up." $\sum_{k} R_{k}$ means $R_{1} + R_{2} + \dots + R_{n}$ — the subscript $k$ just walks through each resistor. It saves writing long " $+ \dots +$ " chains.
Sign / polarity convention: on each resistor we draw the current arrow in the direction we assume charge flows, and mark the end the current enters as $+$ and the end it leaves as $-$ (current always drops from higher to lower potential through a resistor). KVL then reads $+$ for a rise (battery) and $-$ for each drop as you loop; KCL counts arrows into a node as positive. See the figure below.

The three little circuits above are the mental pictures every question on this page is testing — glance back at them whenever a "picture" feels shaky. Notice: in the series sketch one arrow threads through both resistors (same $I$ ); in the parallel sketch the arrow splits at the top node and rejoins at the bottom (same $V$ , currents add); the mixed sketch shows $R_{1}$ in series with a parallel pair.

True or false — justify

Adding more resistors in series always increases the total resistance.

True. Series means one path, so each extra obstacle piles on more voltage drop for the same current —

R_{e q} = \sum_{k} R_{k}

can only grow.

Adding more resistors in parallel always decreases the total resistance.

True. Each new branch is an extra road for current, so the conductance

\sum_{k} G_{k} = \sum_{k} 1/ R_{k}

grows, meaning

R_{e q}

shrinks — below even the smallest branch.

In a parallel combination the equivalent resistance can equal the largest branch resistance.

False.

R_{e q}

is always strictly smaller than the smallest branch, since adding any second path only opens more room for current.

Two identical resistors give the same

R_{e q}

whether wired in series or parallel.

False. Series gives

2 R

, parallel gives

R /2

— a factor of 4 apart. Same parts, opposite wiring, opposite effect.

In series, the biggest resistor has the biggest voltage across it.

True. Current

I

is shared, and

V_{k} = I R_{k}

, so voltage scales directly with

R_{k}

— the biggest obstacle gets the biggest share of the push.

In parallel, the biggest resistor carries the biggest current.

False. Voltage is shared, and

I_{k} = V / R_{k}

, so current is inversely proportional to

R

— the smallest resistor grabs the most current.

If you double every resistance in a network,

R_{e q}

also doubles.

True. Both

\sum_{k} R_{k}

(series) and

(\sum_{k} 1/ R_{k})^{- 1}

(parallel) scale linearly with a common factor, so nesting them still scales

R_{e q}

by that factor.

A wire (zero resistance) placed in parallel with a resistor makes the pair behave like the resistor.

False. It short-circuits the resistor:

R_{e q} = R \cdot 0/ (R + 0) = 0

. All current takes the free path and the resistor is bypassed.

A break (infinite resistance) in one branch of a parallel pair leaves the other branch working normally.

True. The broken branch simply carries no current; the surviving branch still sees the same voltage across the same nodes and behaves as if alone.

Spot the error

"For

3 Ω

and

6 Ω

in parallel,

R_{e q} = 3 + 6 = 9 Ω

Wrong — that's the series answer. Parallel adds conductances (

G = 1/ R

R_{e q} = \frac{3 \cdot 6}{9} = 2 Ω

, which must be below

3 Ω

"I wrote

\frac{1}{R _{e q}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}

, so

R_{e q} = \frac{1}{2} Ω

The reciprocal was skipped:

\frac{1}{2}

1/ R_{e q}

(the total conductance), so you must flip it —

R_{e q} = 2 Ω

"The two wires look like one straight line, so those resistors are in series."

Looking straight isn't enough. Check for a node where current splits; if it does, they're parallel. Only an unbranched path is truly series.

"Both parallel branches share the same voltage, so they must carry equal current."

Same voltage does not mean same current unless the resistances are equal.

I_{k} = V / R_{k}

, so unequal

R

gives unequal current.

n

equal resistors

R

in parallel give

n R

Backwards. Parallel of

n

equal resistors gives

R / n

(more paths, less resistance).

n R

is the series result.

"To find total current from the battery I add the branch currents I haven't computed yet."

Circular. First collapse to

R_{e q}

and use

I = V / R_{e q}

from Ohm's Law; branch currents follow from the shared node voltage afterward.

"The battery's internal resistance is in parallel with the external circuit."

No — internal resistance sits in series with the external load (see EMF and Internal Resistance); the same current flows through it and the load.

Why questions

Why is the current the same through every resistor in series?

There is only one path, and charge cannot accumulate at any point (KCL), so whatever enters the first resistor must exit the last.

Why is the voltage the same across every resistor in parallel?

Voltage is a difference between two nodes, and all parallel branches connect the exact same two nodes — so each feels the identical difference.

Why do we add conductance

G = 1/ R

in parallel rather than

R

Because currents add (KCL):

I = \sum_{k} V / R_{k} = V \sum_{k} 1/ R_{k}

. The quantity that sums is

1/ R

(conductance), and

I / V = 1/ R_{e q}

Why does the "product over sum" shortcut

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

work for exactly two resistors?

It's just

(\frac{1}{R _{1}} + \frac{1}{R _{2}})^{- 1} = (\frac{R _{2} + R _{1}}{R _{1} R _{2}})^{- 1}

rewritten. For three or more you cannot collapse it to a single fraction this cleanly.

Why must these formulas come only from Ohm's law plus Kirchhoff's laws?

Ohm's law relates each resistor's

V

and

I

; KVL fixes how voltages combine in series; KCL fixes how currents combine in parallel. Nothing else is assumed (see Kirchhoff's Laws).

Why does a mixed network get collapsed innermost-first?

You can only apply a series or parallel rule to a group that is purely one or the other. Reducing the innermost pure group turns a mess into a simpler network you can classify again.

Edge cases

A single resistor

R

"in series with nothing else" — what is

R_{e q}

Just

R

itself. A chain of one term sums to that one term; there is no other drop to add.

A single resistor

R

"in parallel with nothing else" — what is

R_{e q}

Also just

R

. The reciprocal sum has one term,

1/ R_{e q} = 1/ R

, so it flips straight back to

R

— the symmetric trivial case of the series one.

One resistor of a series chain is replaced by a plain wire (

0 Ω

). What happens?

R_{e q}

drops by that resistor's value; the current rises and that segment shows zero voltage across it — but the rest of the chain still limits the current.

One resistor of a series chain becomes an open break (

\infty

). What is the current?

Zero everywhere. Series has a single path, so one break stops all flow; every working resistor now shows no voltage drop.

Two equal resistors

R

in parallel — what is

R_{e q}

, and how does current split?

R_{e q} = R /2

, and the current splits exactly in half because equal resistance means equal

I_{k} = V / R

As one parallel branch's resistance

R_{1} \to \infty

, what does

R_{e q}

approach?

It approaches the other branch's resistance

R_{2}

. An infinitely resistive branch carries no current and drops out, leaving only

R_{2}

As one parallel branch's resistance

R_{1} \to 0

, what does

R_{e q}

approach?

It approaches

0

. A near-perfect conductor short-circuits the pair;

\frac{0 \cdot R _{2}}{0 + R _{2}} = 0

, and nearly all current takes that branch.

A network of resistors that is neither purely series nor purely parallel — can these two rules always solve it?

No. A balanced or unbalanced bridge (see Wheatstone Bridge) has no pure series/parallel groups; you need full Kirchhoff analysis instead.

Connections

Ohm's Law — $V = I R$ , invoked in every branch above.
Kirchhoff's Laws — KVL justifies the "same current / voltages add" trap; KCL justifies the "same voltage / currents add" trap.
EMF and Internal Resistance — the series-vs-parallel internal-resistance trap.
Wheatstone Bridge — the edge case where these rules run out.
Power in Circuits — a natural follow-up once currents/voltages are known.
Parent: Hinglish topic note.