Resistors in parallel

Flip on a fan, a light and a phone charger in the same room and every one of them works at full strength, independent of the others — switch the fan off and the light does not even flicker. That everyday convenience is the signature of a parallel combination of resistors, the second of the two basic ways to wire resistors together (after series) in the CBSE Class 10 chapter "Electricity". In a parallel circuit, resistors sit side by side between the same two points, each forming its own branch for current, and the two combinations are almost always compared and contrasted on the exam. This article builds the defining rule of a parallel circuit, derives the equivalent-resistance formula from Ohm's law, and applies it to household wiring and to fully worked numericals.

The core insight

In a parallel combination every resistor shares the same potential difference, so the currents through the branches add up while the reciprocals of the resistances add up to give a smaller equivalent resistance.

The core idea

Resistors are in parallel when they are connected between the same pair of points, so each one forms a separate branch and current has multiple alternative paths between those two points. The defining electrical feature is that the same potential difference $V$ exists across every resistor, since all branches connect the identical two points — this is the opposite of series, where current is common and voltage divides. Because each branch has its own resistance, each draws its own share of current from the source, and the total current equals the sum of the branch currents. For resistors $R_1$ , $R_2$ , $R_3$ in parallel across a battery of potential difference $V$ , with branch currents $I_1$ , $I_2$ , $I_3$ and total current $I$ :

I = I_1 + I_2 + I_3

where: $I$ is the total current drawn from the source, and $I_1$ , $I_2$ , $I_3$ are the currents through each individual branch. Applying Ohm's law separately to each branch, using the same $V$ throughout, and to the combination as a whole (with $R_p$ the equivalent resistance), leads to the parallel-resistance rule:

\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dots

where: $R_p$ is the equivalent (total) resistance of the combination, and $R_1$ , $R_2$ , $R_3, \dots$ are the individual resistances of each branch. A key consequence: $R_p$ is always smaller than the smallest individual resistance in the combination. Adding more parallel branches opens more paths for current, so it always lowers the total resistance and raises the total current drawn from the source — the reverse of what happens when resistors are added in series.

Figure

Three resistors connected side by side between the same two points across a battery, each carrying its own branch current, with the currents adding up at the battery terminal

Everyday application — household wiring. Homes wire all appliances and light points in parallel, for two reasons. Every appliance receives the full mains voltage directly, no matter how many other appliances are switched on, so a fan and a bulb both work at their rated voltage. Each appliance can also be switched on or off independently — in a series wiring, switching off one appliance would break the circuit for everything else on that line.

Derivation AEquivalent resistance of resistors in parallel

The goal is to show, starting only from the definition of a parallel connection (same potential difference across every branch) and Ohm's law applied branch by branch, that the reciprocal of the equivalent resistance of a parallel combination equals the sum of the reciprocals of the individual resistances — the formula used in every parallel-circuit numerical.

Step 1. Set up the branches. Let $R_1$ , $R_2$ , $R_3$ be connected between the same two points, across a source of potential difference $V$ . Because all three share the same two points, the potential difference across each resistor is the same value $V$ .

Step 2. Apply Ohm's law to each branch separately. The current through each resistor is $I_1 = V/R_1$ , $I_2 = V/R_2$ , $I_3 = V/R_3$ — each using the identical $V$ , since voltage does not divide in parallel the way it does in series.

Step 3. Apply Ohm's law to the whole combination. If $R_p$ is the single equivalent resistance that would draw the same total current $I$ from the same source $V$ , then $I = V/R_p$ . Since the total current is the sum of the branch currents, $I = I_1 + I_2 + I_3$ , substituting gives $\dfrac{V}{R_p} = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dfrac{V}{R_3}$ .

Step 4. Divide through by the common $V$ . Since $V$ appears in every term, it cancels completely, leaving $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$ , which generalises to any number of resistors in parallel. Because reciprocals are being added, $R_p$ always comes out smaller than the smallest $R$ in the set — a quick sanity check on any final numerical answer.

Worked example

Two resistors of 4 Ω and 6 Ω are connected in parallel across a 12 V battery. Find (a) the equivalent resistance, (b) the current through each resistor, and (c) the total current drawn from the battery.

Given: $R_1 = 4\ \Omega$ , $R_2 = 6\ \Omega$ , $V = 12\ \text{V}$ .

Step 1. Find the equivalent resistance. $\dfrac{1}{R_p} = \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12}$ , so $R_p = \dfrac{12}{5} = 2.4\ \Omega$ .

Step 2. Find the branch currents. Since the same $V = 12$ V acts across each resistor in parallel, $I_1 = \dfrac{12}{4} = 3\ \text{A}$ and $I_2 = \dfrac{12}{6} = 2\ \text{A}$ .

Step 3. Find the total current. $I = I_1 + I_2 = 3\ \text{A} + 2\ \text{A} = 5\ \text{A}$ . As a check, $I = V/R_p = 12/2.4 = 5\ \text{A}$ , which matches.

Answer: equivalent resistance = 2.4 Ω, branch currents are 3 A and 2 A, and the total current drawn is 5 A.

Competitive stretch

A 60 W bulb and a 100 W bulb, both rated 220 V, are connected in parallel to a 220 V mains supply. Since parallel wiring gives each bulb the full 220 V, both glow at their rated brightness regardless of the other. Using $I = P/V$ at the same voltage, the 100 W bulb draws $I = 100/220\ \text{A}$ and the 60 W bulb draws $I = 60/220\ \text{A}$ , so the higher-wattage bulb draws more current — power and current rise together when voltage is fixed.

Common traps

Forgetting to take the reciprocal at the end — computing $\frac{1}{R_1} + \frac{1}{R_2}$ correctly but reporting that sum itself as $R_p$ , instead of inverting it.
Assuming current is the same across every resistor in parallel — it is voltage that stays the same; current divides unequally, with more current through the smaller resistance.
Forgetting that $R_p$ in a parallel combination is always smaller than any individual resistor, a useful check on numerical answers.
Believing household circuits use series wiring "for safety" — parallel wiring is used so each appliance works independently at full voltage, and one failure does not disable the rest.

Ask the tutor: Want the AI tutor to work through a series and a parallel combination of the same resistor set side by side, so the contrast between them is crystal clear?

Why is the equivalent resistance in parallel always smaller than the smallest individual resistor?

Adding a resistor in parallel opens up an extra path for current to flow between the same two points, so for the same applied voltage more total current can flow than through any single branch alone. Since resistance is voltage divided by current, more current at the same voltage means less overall resistance. Mathematically, $\frac{1}{R_p}$ is a sum of positive reciprocals, so it is always larger than any single $\frac{1}{R}$ term, which forces $R_p$ itself to be smaller than the smallest resistor in the set.

If I add a third resistor in parallel, does the total current from the battery increase or decrease?

It increases. Adding another branch does not change the voltage across the existing resistors — that stays fixed at the source voltage $V$ — but it does add another current path, so the total current $I = I_1 + I_2 + I_3$ picks up one more term. At the same time the equivalent resistance $R_p$ drops further, consistent with $I = V/R_p$ rising as $R_p$ falls. This is exactly why plugging in more appliances at home increases the total current drawn from the mains.

How is a parallel combination different from a series combination in terms of what stays constant?

In series, the resistors share one single path, so the current is forced to be the same through every resistor, while the total voltage splits unequally across them depending on resistance. In parallel, the resistors sit on separate branches between the same two points, so the voltage is the same across every resistor, while the total current splits unequally across the branches. The equivalent resistance also behaves oppositely: it is always larger than the biggest resistor in series, but always smaller than the smallest resistor in parallel.

Resistors in parallel

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads