Question

Which of the following circuits has the smallest equivalent resistance?The value of the last resistance in the fourth circuit is 2 ohm.

Answer 1

Given:

Four circuit diagrams of the resistance are shown in the question.

To find:

The circuit with the smallest equivalent resistance

Explanation:

The equivalent resistance in the first circuit is,

$\begin{gathered} 2+2\text{ \lparen series combination\rparen} \\ =4\text{ ohm } \end{gathered}$

The equivalent resistance of the second circuit is,

$\begin{gathered} \frac{1}{R}=\frac{1}{2}+\frac{1}{2}\text{ \lparen parallel combination\rparen} \\ \frac{1}{R}=\frac{2}{2} \\ R=1\text{ ohm} \end{gathered}$

The equivalent resistance of the third circuit is,

$\begin{gathered} 2+2+2+2\text{ \lparen series\rparen} \\ =8\text{ ohm} \end{gathered}$

The equivalent resistance of the third circuit is,

$\begin{gathered} \frac{1}{R^{\prime}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2} \\ \frac{1}{R^{\prime}}=\frac{4}{2} \\ R^{\prime}=\frac{2}{4} \\ R^{\prime}=\frac{1}{2}\text{ ohm} \end{gathered}$

Hence, the circuit with the smallest resistance is the fourth circuit where four resistances are in parallel combination.