Question

Three resistors are connected using wires of negligible resistance. What is the approximate resistance between the two points P and Q?

Answer 1

Hint:Equivalent resistance between two points can be evaluated after simplifying the circuit into simple series parallel combinations of resistors and then applying the appropriate formula. When the resistors are connected in series their equivalent is given by ${R_s} = {R_1} + {R_2}........ + {R_n}$ . When the resistors are connected in parallel, the equivalent resistance is given by $\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ......... + \dfrac{1}{{{R_n}}}$.

Complete step by step answer:
A series connection is identified to be the connection where the resistors are connected in a chain like fashion whereas in a parallel connection the resistors have the same starting and ending points in the circuit. In a simple conductor, when there is no resistance, there is no potential drop and hence the potential difference remains zero. This means that along a straight conductor, the potential remains the same. The circuit given to us is,

On inspection, the circuit reduces to

We can clearly see that the three resistors are in parallel with each other.
Let the equivalent resistance between points P and Q be ${R_{eq}}$ .
Using the formula for the parallel connection of resistors,
$\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ......... + \dfrac{1}{{{R_n}}}$
Here we have three resistors so this formula reduces to $\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$ .
Let ${R_1} = 1.0\,\Omega $ , ${R_2} = 2.0\,\Omega $ and ${R_3} = 3.0\,\Omega $ .
Substituting these values in the formula,
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3}$
Further solving the equation, we get,
$\dfrac{1}{{{R_{eq}}}} = \dfrac{{6 + 3 + 2}}{6}$
$ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{{11}}{6}$
Hence, ${R_{eq}} = \dfrac{{11}}{6}\,\Omega $ .

So, the equivalent resistance between the two points P and Q is ${R_{eq}} = \dfrac{{11}}{6}\,\Omega $.

Note:We must carefully analyze the circuit for the series and parallel combinations. Sometimes it might happen due to the diagrammatic representation, we may perceive a wrong combination. Always remember that in a series combination, there is a voltage drop across all the resistors and the current remains the same. However, if the current gets distributed and there is no potential drop then it is a parallel combination.