Question

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be
$A)1:2$
$B)2:1$
$C)1:4$
$D)4:1$

Answer 1

First, the equivalent resistances have to be calculated for both series and parallel connections. Note that, there is no effect of lengths and diameters on the resistance since they are equal.
Next, the relationship between produced heat and resistance is to be used for two cases and find the ratio of produced heat.

Formula used:
Let the resistances for two conducting wires are ${R_1}$ and ${R_2}$.
So, equivalent resistance in series connection ${R_s} = {R_1} + {R_2}$
And, equivalent resistance in parallel connection ${R_p} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}$
The relation between produced heat $(H)$ and resistance $(R)$ is, $H = \dfrac{{{V^2}t}}{R}$
Where, $V$ is the potential difference and $t$ is the time.

Complete step-by-step solution:
Two conducting wires of the same material, same lengths and, same diameters are connected in two ways.
First in series and second parallel connection.
Let the resistances for two conducting wires are ${R_1}$ and ${R_2}$.
Since, the material, lengths and diameters are the same the resistances are equal. Hence, ${R_1} = {R_2} = R$
So, equivalent resistance in series connection ${R_s} = {R_1} + {R_2} = 2R$
And, equivalent resistance in parallel connection ${R_p} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{{R^2}}}{{2R}} = \dfrac{R}{2}$
The relation between produced heat $(H)$ and resistance $(R)$ is, $H = \dfrac{{{V^2}t}}{R}$
Where, $V$ is the potential difference and $t$ is the time.
So, $H \propto \dfrac{1}{R}$
$\therefore \dfrac{{{H_s}}}{{{H_p}}} = \dfrac{{{R_p}}}{{{R_s}}}$ where, ${H_s}$ and ${H_p}$ heat produced in series and parallel combinations respectively.
$\Rightarrow \dfrac{{{H_s}}}{{{H_p}}} = \dfrac{{R/2}}{{2R}}$
$\Rightarrow \dfrac{{{H_s}}}{{{H_p}}} = \dfrac{1}{4}$
So the ratio of ${H_s}$ and ${H_p}$ is $1:4$ .
Hence, the right Option is $C) \Rightarrow 1:4$

Note: For a conductor of length $l$, area $A$ and, electric conductance $\rho$, the resistance $R = \rho \dfrac{l}{A}$
Hence for two conductors of the same material, same lengths and, same diameters, the resistances will be the same.