Question

A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio ${\displaystyle \frac{R}{R^{\prime }}}$ is
$\begin{array}{rl}& A){\displaystyle \frac{1}{25}}\\ & B){\displaystyle \frac{1}{5}}\\ & C)5\\ & D)25\end{array}$

Answer 1

Hint: Firstly we know the wire of resistance R is cut into 5 equal parts we will find the resistance of each part then we will connect them in parallel and find equivalent resistance R’. After that, we will take the ratio and obtain the answer.

Complete step by step answer:
We know that,
 R (resistance) is directly proportional to Length L.
When the wire is cut into 5 equal pieces, its resistance will become = ${\displaystyle \frac{1}{5}}th$
The resistance of each piece of wire after cut into 5 pieces will be given as ${\displaystyle \frac{R}{5}}\mathrm{\Omega }$
Now, these wires are connected in parallel.

We know that in parallel combination equivalent resistance of the circuit is given by:

${\displaystyle \frac{1}{R^{\prime }}}={\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}+{\displaystyle \frac{1}{R_3}}+{\displaystyle \frac{1}{R_4}}+{\displaystyle \frac{1}{R_5}}$

$\begin{array}{rl}& {\displaystyle \frac{1}{R^{\prime }}}={\displaystyle \frac{1}{{\displaystyle \frac{R}{5}}}}+{\displaystyle \frac{1}{{\displaystyle \frac{R}{5}}}}+{\displaystyle \frac{1}{{\displaystyle \frac{R}{5}}}}+{\displaystyle \frac{1}{{\displaystyle \frac{R}{5}}}}+{\displaystyle \frac{1}{{\displaystyle \frac{R}{5}}}}\\ & {\displaystyle \frac{1}{R^{\prime }}}={\displaystyle \frac{5}{R}}+{\displaystyle \frac{5}{R}}+{\displaystyle \frac{5}{R}}+{\displaystyle \frac{5}{R}}+{\displaystyle \frac{5}{R}}\\ & {\displaystyle \frac{1}{R^{\prime }}}={\displaystyle \frac{25}{R}}\\ & R^{\prime }={\displaystyle \frac{R}{25}}\end{array}$
Now we will find the ratio ${\displaystyle \frac{R}{R^{\prime }}}$

$\begin{array}{rl}& {\displaystyle \frac{R}{R^{\prime }}}={\displaystyle \frac{R}{{\displaystyle \frac{R}{25}}}}\\ & {\displaystyle \frac{R}{R^{\prime }}}={\displaystyle \frac{25R}{R}}\\ & {\displaystyle \frac{R}{R^{\prime }}}=25\end{array}$

So from the above calculations, we obtain the correct answer is (D).

Additional Information
Advantages and disadvantages of parallel connection of resistors:
Parallel circuits provide an equivalent voltage to each source and appliance within the circuit, thus all appliances work efficiently. In a parallel circuit, if one appliance is fused, the current continues to flow through the others.
The total resistance of a circuit is adequate to the sum of individual resistances. The voltage applied to a circuit is adequate to the sum of the individual voltage drops. The drop across a resistor during a circuit is directly proportional to the dimensions of the resistor.

Note:
In the series resistor network we know that the total resistance of the circuit was equal to the sum of all the individual resistors added together. For resistors in parallel, the equivalent circuit resistance is calculated differently.
for finding the equivalent resistance the reciprocal value of the individual resistances are added together instead of the resistances themselves with the inverse of the algebraic sum.