Question

Ten identical wires each having a resistance of one ohm are connected in parallel. The combination will have a resistance of
(A)$0.1\;{\rm{\Omega }}$
(B)$10\;\Omega$
(C)$0.01\;{\rm{\Omega }}$
(D)$1\;{\rm{\Omega }}$

Answer 1

Resistance of a wire is defined as ability of the wire to oppose the flow of current and it depends on the dimension, resistivity and the physical properties of the material. The equivalent resistance is the combined effect of all of the resistance that are connected in a circuit. If the resistances are connected in parallel, then we can calculate the equivalent resistance by the given formula:
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}.......$
Here, ${R_{eq}}$ is the equivalent resistance, ${R_1}$ ,${R_2}$ … are the resistance of wires.

Complete step by step answer:
The resistance of the wire is: $R = 1\;{\rm{\Omega }}$
The number of wire is: $10$
It is given that the wires are connected in the parallel connection.
The equivalent resistance of the ten wires connected in parallel connection is given as follows,
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ..... + \dfrac{1}{{{R_{10}}}}$
Here, all the wires have the same resistance value so the value of ${R_1}$, ${R_2}$ ….${R_{10}}$ is the same and that is equal to the one ohm.
Substitute all the values in the above expression,
$\begin{array}{l} \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{1}\\\ \dfrac{1}{{{R_{eq}}}} = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1\\\ \dfrac{1}{{{R_{eq}}}} = 10 \end{array}$
Evaluate further
$\begin{array}{l} \dfrac{1}{{{R_{eq}}}} = 10\\\ {R_{eq}} = \dfrac{1}{{10}}\\\ {R_{eq}} = 0.1\;{\rm{\Omega }} \end{array}$
Therefore, the combination of the resistance has the value of $0.1\;{\rm{\Omega }}$ and the correct answer is the ${\rm{option}}\;{\rm{(A)}}$.

Note: : In such types of questions If the resistances are in series connection, then the equivalent resistance can be calculated by just taking the summation of the resistances and the value of current remains same in that connection. The formula for equivalent resistance in the series connection is given as follows,
${R_{eq}} = {R_1} + {R_2} + .......$
Addition Information: If there is loop that has some resistance $R$ and the flow of the current is $I$, then the voltage (V) or potential difference will be proportional to the current in that connection and the value of the voltage will be the product of resistance(R) and the current(I).