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

The correct option is C

$1:4$

Step 1: Given parameters

Two conducting wires, are made of the same material and have equal lengths and diameters.

That means they have the same resistance $R$.

Step 2: Calculate the resistance and power in series combination.
For two resistance in series Combination,

$\begin{array}{rcl}{R}_{series}& =& R_1+R_2\\ & =& R+R\\ & =& 2R\end{array}$
Now the power is given by,

$\begin{array}{rcl}{P}_{series}& =& \frac{V^2}{R_{series}}\\ & =& \frac{V^2}{2R}.........\left(1\right)\end{array}$

Step 3: Calculate the resistance and power in parallel combination.

For two resistance in parallel Combination,

$\begin{array}{rcl}\frac{1}{R_{parallel}}& =& \frac{1}{R_1}+\frac{1}{R_2}\\ & =& \frac{1}{R}+\frac{1}{R}\\ & =& \frac{2}{R}\end{array}$

So we can write,$R_{parallel}=\frac{R}{2}$.

Now the power is given by,

$\begin{array}{rcl}{P}_{parallel}& =& \frac{V^2}{R_{parallel}}\\ & =& \frac{V^2}{{\displaystyle \frac{R}{2}}}\\ & =& \frac{2V^2}{R}.........\left(2\right)\end{array}$

Step 4: Find the ratio of the heat produced in series and parallel combinations.

Now the ratio of heat produced in series and parallel combinations can be given by dividing, (2) from (1), i.e.,

$\begin{array}{rcl}\frac{P_{series}}{P_{parallel}}& =& \frac{{\displaystyle \frac{V^2}{2R}}}{{\displaystyle \frac{2V^2}{R}}}\\ & =& \frac{1}{4}\end{array}$

Hence, the ratio of heat produced in series and the parallel combination is $\frac{1}{4}$.

Therefore, option (C) is correct.