Series and parallel combination of resistors — find equivalent resistance

Question

Three resistors of $2\Omega$ , $3\Omega$ , and $6\Omega$ are connected in (a) series and (b) parallel. Find the equivalent resistance in each case. A 12V battery is connected — find the current drawn in each case.

(NCERT Class 10 — Electricity)

Solution — Step by Step

Part (a): Series Combination

In series, resistances simply add up:

R_{eq} = R_1 + R_2 + R_3 = 2 + 3 + 6 = \mathbf{11\,\Omega}

I = \frac{V}{R_{eq}} = \frac{12}{11} = \mathbf{1.09\text{ A}}

In series, this same current flows through all three resistors.

Part (b): Parallel Combination

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}

= \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1

R_{eq} = \mathbf{1\,\Omega}

I = \frac{V}{R_{eq}} = \frac{12}{1} = \mathbf{12\text{ A}}

In parallel, the voltage across each resistor is the same (12V), but the current splits. Individual currents: $I_1 = 12/2 = 6$ A, $I_2 = 12/3 = 4$ A, $I_3 = 12/6 = 2$ A. Total = 12 A. Checks out.

Why This Works

Series: Current has only one path, so the same current flows through each resistor. Each resistor drops some voltage, and the total voltage equals the sum of drops: $V = IR_1 + IR_2 + IR_3 = I(R_1 + R_2 + R_3)$ .

Parallel: Each resistor gets the full battery voltage. Each provides a separate path for current. The total current is the sum of individual currents: $I = V/R_1 + V/R_2 + V/R_3$ .

Key pattern: series always gives a larger equivalent resistance than the largest individual resistor. Parallel always gives a smaller equivalent resistance than the smallest individual resistor.

Alternative Method

For two resistors in parallel, use the shortcut: $R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}$ . For our three resistors, combine two first: $R_{2,3} = \frac{3 \times 6}{3 + 6} = 2\Omega$ . Then combine with the third: $R_{eq} = \frac{2 \times 2}{2 + 2} = 1\Omega$ . Same answer.

Quick check for parallel combinations: the equivalent resistance must be less than the smallest resistor. Here, the smallest is $2\Omega$ and we got $1\Omega$ . If your answer is greater than $2\Omega$ for a parallel combination, something is wrong.

Common Mistake

In parallel, students often forget to take the reciprocal at the end. They calculate $\frac{1}{R_{eq}} = 1$ and write $R_{eq} = 1$ . In this problem, it happens to be correct since $1/1 = 1$ . But if $\frac{1}{R_{eq}} = 0.5$ , the answer is $R_{eq} = 2\Omega$ , not $0.5\Omega$ . Always remember to flip the final reciprocal.