Question

Two conducting wires of the same material and of equal lengths and equal diameter 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?

One who will answer correctly and briefly awarded by 50 points and mark as __________ ( you know ).
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Answer 1

The wires are of same material, of equal lengths and equal diameters. Hence, their resistance will be equal.

When connected in series, $\sf R_{eq} = R + R$

$\sf R_{eq} = Equivalent\ Resistance$

So in series, total resistance = 2R

In parallel, $\sf \frac{1}{R_{eq}}= \frac{1}{R}+\frac{1}{R}$

$\sf\implies R_{eq} = \frac{R}{2}$

Potential difference, V, is same in both the cases

Heat produced in time 't' = $\sf \frac{V^2}{R}t$

In first case, heat produced = $\sf\frac{V^2}{(2R)}t$

In second case, heat produced = $\sf\frac{V^2}{(\frac{R}{2})^2}t$

$\sf\implies \frac{4V^2}{R}t$

So, ratio of them = 1 : 4

Ratio of heat produced in series and parallel combinations = 1 : 4

Answer 2

GIVEN :-

Two conducting wires of the same material and of equal lengths and equal diameter are first connected in series and then parallel in a circuit across the same potential difference.

TO FIND :-

The ratio of heat produced in series and parallel combinations.

SOLUTION :

Equivalent resistance when they are connected in series :-

Req = R + R = 2R

Let R' = 2R

$\rule{200}{2}$

Equivalent resistance when they are connected in parallel :-

1/Req = 1/R + 1/R = 2/R

⇒Req = R/2

Let R" = R/2

$\rule{200}{2}$

Now, we know,

$\boxed{\sf{Heat\ energy=\frac{I^2}{R}t }}$

Ratio of the heat produced in series and parallel combinations :-

$\sf{\dfrac{I^2}{R'}t \div \dfrac{I^2}{R''}t}\\\\\\\sf{=\dfrac{I^2t}{R'} \times \dfrac{R''}{I^2t}}\\\\\\\sf{=\dfrac{R''}{R} }\\\\\\\sf{=\dfrac{\dfrac{R}{2} }{2R} }\\\\\\\sf{=\dfrac{R}{2}\times \dfrac{1}{2R} }\\\\\\\sf{=\dfrac{1}{4} }$

So, the ratio the heat produced in series and parallel combinations is 1 : 4.