For n equal resistors in parallel: R_eq = R/n = 6/3 = 2Ω. Full derivation from 1/R formula.

Three 6Ω Resistors in Parallel — Find Equivalent Resistance

Question

Three resistors, each of resistance 6Ω, are connected in parallel. Find the equivalent resistance of the combination.

Solution — Step by Step

For resistors in parallel, the reciprocals add up:

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

We use reciprocals because parallel resistors give current more paths to flow — more paths means less overall opposition.

All three resistors are 6Ω, so:

\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6}

\frac{1}{R_{eq}} = \frac{3}{6} = \frac{1}{2}

R_{eq} = 2\ \Omega

Taking the reciprocal of both sides gives us the final answer directly.

Equivalent resistance = 2Ω

For n equal resistors of resistance R each, all in parallel, the shortcut is:

R_{eq} = \frac{R}{n}

Here, $R_{eq} = \frac{6}{3} = 2\ \Omega$ . Memorise this — it saves 30 seconds in board exams.

Why This Works

When resistors are in parallel, every resistor connects directly between the same two points (same potential difference across each). So the total current from the battery splits up and flows through each branch independently.

Adding more parallel resistors is like opening more gates — the overall resistance drops because the circuit offers the current more routes. That’s why $R_{eq}$ in parallel is always less than the smallest individual resistor.

Here, each branch takes $\frac{V}{6}$ amperes. Three identical branches together draw $3 \times \frac{V}{6} = \frac{V}{2}$ amperes total. By Ohm’s law, a current of $\frac{V}{2}$ from a voltage $V$ means the equivalent resistance is $\frac{V}{V/2} = 2\ \Omega$ . This is exactly the algebraic result — the two approaches agree perfectly.

Alternative Method

Use the general formula directly for n equal resistors:

R_{eq} = \frac{R}{n} = \frac{6\ \Omega}{3} = \mathbf{2\ \Omega}

This one-line method works only when all resistors are equal. For mixed values you must go back to the $\frac{1}{R_{eq}}$ formula. In NCERT numericals, equal-resistor problems appear often enough that this shortcut is worth keeping handy.

Common Mistake

Forgetting to take the reciprocal at the end.

Students correctly calculate $\frac{1}{R_{eq}} = \frac{1}{2}$ and then write $R_{eq} = \frac{1}{2}\ \Omega = 0.5\ \Omega$ as the answer — forgetting that the formula gives $\frac{1}{R_{eq}}$ , not $R_{eq}$ itself.

Always check: your final $R_{eq}$ must be less than the smallest resistor in the combination. If you got 0.5Ω when the smallest resistor is 6Ω, something is right. But if you ever get a value larger than any individual resistor after a parallel calculation, you’ve made this exact error.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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