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

The heat formula is: $Heat = I^2Rt$ where I is current, R is resistance and t is time.

To take ratios of heat in both series and parallel connection we are taking the same resistance R, the same current I.

So we have to take the same time t as well. Then only we can compare the heats in series and parallel combination.

When the wires are connected in series, then Resistance $=R+R=2R$

The heat produced $H_{series}=I^2(2R)t=2I^2Rt$

When the two wires of resistance $R$ are connected in parallel, then resistance $=\frac{1}{\frac{1}{R}+\frac{1}{R}}$

$=\frac{1}{\frac{2}{R}}$

$=\frac{R}{2}$

Then heat produced $H_{parallel}=I^2(\frac{R}{2})t$

$=\frac{1}{2}I^2Rt$

So, $\frac{H_{series}}{H_{parallel}}=\frac{2I^2Rt}{\frac{1}{2}I^2Rt}$

$=\frac{4}{1}$

Or $H_{series}:H_{parallel}=4:1$

So, option $(d)$ is correct.

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Updated on: 10-Oct-2022

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