Two resistors of resistance, 100 Ω and 200 Ω are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 Ω to that in 200 Ω in a given time is :

Question

Two resistors of resistance, 100 Ω and 200 Ω are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 Ω to that in 200 Ω in a given time is : (A) 4: 1 (B) 1: 2 (C) 2: 1 (D) 1: 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/34ce1e2ba659cb4c6265bb9fffe28bab-.png" /> As both resistors are in parallel combination so potential drop ( V ) across both are same. P =( V 2/ R ) ⇒ P propto (1/ R ) ( P 1/ P 2)=( R 2/ R 1)=(200/100)=(2/1) =2: 1

Q. Two resistors of resistance, $100 Ω$ and $200 Ω$ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in $100 Ω$ to that in $200 Ω$ in a given time is :

$4 : 1$

$1 : 2$

$2 : 1$

$1 : 4$

As both resistors are in parallel combination so potential drop $(V)$ across both are same.
$P = \frac{V ^{2}}{R} \Rightarrow P \propto \frac{1}{R}$
$\frac{P _{1}}{P _{2}} = \frac{R _{2}}{R _{1}} = \frac{200}{100} = \frac{2}{1}$
$= 2 : 1$

Q. Two resistors of resistance, 100Ω and 200Ω are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100Ω to that in 200Ω in a given time is :

4:1

1:2

2:1

1:4