Prove that when a current is divided between two resistors in accordance with Kirchhoff’s laws, the heat produced is minimum.

Solution

Consider two resistors R₁ and R₂ connected in parallel and the current through the various arms of the circuit be as shown in the figure below.

According to Kirchhoff’s first law, at junction A,

i = i₁ + i₂ or i₂ = i – i₁ ...(i)

Let H be the heat produced in the circuit in t seconds, then

H = i₁²R₁t + i₂²R₂t
= i₁²R₁t + (i – i₁)² R₂t [from (i)]

In case the heat produced in the circuit is minimum, then

$\frac{dH}{{di}_{i}} = 0;$

therefore,

2i₁R₁t + 2(i – i₁) (–1) R₂t = 0
2i₁R₁t – 2i₂ R₂t = 0
i₁R₁ – i₂R₂ = 0

which is according to Kirchhoff’s second law in a closed circuit ACDEFA.

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Current Electricity

Prove that when a current is divided between two resistors in accordance with Kirchhoff’s laws, the heat produced is minimum.

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