Question

In the circuit shown in the figure, the power developed across $ 1\Omega ,2\Omega $ and $ 3\Omega $ resistances are in the ratio of

Answer 1

Hint: to solve this question we have to know what the resistance is. Resistance is the measure of the opposition to current flow in an electrical circuit. We know that the unit of resistance is ohm. We know that the symbol of ohm is like a Greek word omega. We know that, when resistance increases then the current flow decreases. Because resistance is inversely proportional to current.

Complete step by step answer:
We can see here, the current in parallel divides in inverse ratio of resistance is
 $ {I_1} = (\dfrac{2}{{1 + 2}})i = \dfrac{2}{3}i $
 $ {I_2} = (\dfrac{1}{{1 + 2}})i = \dfrac{1}{3}i $
Here, we can say according to the question, I is the total current. Now, we can consider P is the power and R is the resistance. $ {P_1} $ is the power developed across $ 1\Omega $ resistance. Again, $ {P_2} $ is the power developed across $ 2\Omega $ resistance. Again, $ {P_3} $ is the power developed across the resistance $ 3\Omega $ .
Now, we know that $ P = {I^2}R $
Now we are going to put the values of resistances which are mentioned in the above question.
Now, $ {P_1} = {I_1}^2R = {(\dfrac{2}{3})^2}{i^2} \times 1 = \dfrac{4}{9}{i^2} $
Again, we can write similarly,
 $ {P_2} = {I_2}^2R = {(\dfrac{1}{3})^2}{i^2} \times 2 = \dfrac{2}{4}{i^2} $
Similarly, again, we can write,
 $ {P_3} = {I_3}^2R = {i^2} \times 3 = 3{i^2} $
 Now, the ratio of the three powers are,
 $ {P_1}:{P_2}:{P_3} = \dfrac{4}{3}:\dfrac{2}{9}:3 = 4:2:27 $
This is the right answer.

Note:
We can get confused between the formula of parallel resistance, series resistances and conductors. We also have to remember that when two resistors are parallel then the potential difference becomes common or same. But when two resistors are connected in a series manner then the current flows through all the resistances are the same or constant.