Question

Two heater wires of equal length are first connected in series and then in parallel. The ratio of heat produced in the two cases will be:
(A) $2:1$
(B) $1:2$
(C) $4:1$
(D) $1:4$

Answer 1

Hint: The heater wire is connected in the circuit, if the heat is produced in the wire so we assume that the resistance of the heater wire is $R$. Since, the two wires are of equal length then the resistance of the two wires is the same. By using the resistor in series and resistor in parallel formula, the ratio can be determined.

Useful formula
The resistance in series is given by,
${R_S} = {R_1} + {R_2}$
Where, ${R_s}$ is the total resistance in series, ${R_1}$ and ${R_2}$ are the individual resistance.
The resistance in parallel is given by,
$\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
Where, ${R_P}$ is the total resistance in parallel, ${R_1}$ and ${R_2}$ are the individual resistance.
The heat produced in the resistor is given by,
$H = \dfrac{{{V^2}}}{R}$
Where, $H$ is the heat produced, $V$ is the voltage and $R$ is the resistance.

Complete step by step solution
Given that,
The two heater wires of equal length, then \[{R_1} = {R_2} = R\]
Now,
The resistance in series is given by,
${R_S} = {R_1} + {R_2}$
The two resistance are equal, then the above equation is written as,
${R_S} = R + R$
By adding the terms in the above equation, then the above equation is written as,
${R_S} = 2R\,.................\left( 1 \right)$
Now,
The resistance in parallel is given by,
$\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$
The two resistance are equal, then the above equation is written as,
$\dfrac{1}{{{R_P}}} = \dfrac{1}{R} + \dfrac{1}{R}$
By adding the terms in the above equation, then the above equation is written as,
$\dfrac{1}{{{R_P}}} = \dfrac{2}{R}$
By tasking reciprocal in the above equation, then the above equation is written as,
${R_P} = \dfrac{R}{2}\,.....................\left( 2 \right)$
Now, the heat produced in the series connection is given by,
${H_s} = \dfrac{{{V^2}}}{{{R_s}}}$
By substituting the equation (1) in the above equation, then
${H_s} = \dfrac{{{V^2}}}{{2R}}\,.....................\left( 3 \right)$
Now, the heat produced in the parallel connection is given by,
${H_p} = \dfrac{{{V^2}}}{{{R_p}}}$
By substituting the equation (2) in the above equation, then
${H_p} = \dfrac{{{V^2}}}{{\left( {\dfrac{R}{2}} \right)}}$
By rearranging the terms, then
${H_p} = \dfrac{{2{V^2}}}{R}\,................\left( 4 \right)$
Now, the equation (3) divided by the equation (4), then
$\dfrac{{{H_s}}}{{{H_p}}} = \left( {\dfrac{{\left( {\dfrac{{{V^2}}}{{2R}}} \right)}}{{\left( {\dfrac{{2{V^2}}}{R}} \right)}}} \right)$
By cancelling the same terms, then
$\dfrac{{{H_s}}}{{{H_p}}} = \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{2}{1}} \right)}}$
By rearranging the terms, then
$\dfrac{{{H_s}}}{{{H_p}}} = \dfrac{1}{4}$

Hence, the option (D) is the correct answer.

Note: From the final answer it is very clear that the amount of the heat produced by the resistor in parallel connection is equal to the four times of the heat produced by the resistors in series connection. The heat produced is inversely proportional to the resistor.