Two heating elements with resistances of 100 ohms and 400 ohms are connected in series.

Two heating elements with resistances of 100 ohms and 400 ohms are connected in series. How long will it take in the first element to release the same amount of heat as in the second in 30 minutes?

Since two heating elements with resistance R₁ = 100 Ohm and R₂ = 400 Ohm are connected in series, a current with the same current strength passes through them: I₁ = I₂ = I. The amount of heat that is released on the heating elements during time t can be calculate according to the Joule – Lenz law: Q = I ^ 2 ∙ R ∙ t. To determine how long it takes t₁ in the first element, the same amount of heat Q₁ = Q₂ will be released as in the second in t₂ = 30 minutes, we compose expressions for Q₁ and Q₂ and equate them. We get:
Q₁ = I₁ ^ 2 ∙ R₁ ∙ t₁ and Q₂ = I₂ ^ 2 ∙ R₂ ∙ t₂;
I₁ ^ 2 ∙ R₁ ∙ t₁ = I₂ ^ 2 ∙ R₂ ∙ t₂ or R₁ ∙ t₁ = R₂ ∙ t₂, then t₁ = R₂ ∙ t₂ / R₁. Substitute the values ​​of physical quantities in the calculation formula, we get:
t₁ = 400 Ohm ∙ 30 min / 100 Ohm;
t₁ = 120 minutes = 2 hours.
Answer: in 2 hours.