Question

Two resistors R₁ and R₂ of resistance 3Ω and 6Ω respectively are connected in parallel across a battery of potential difference 12 V. Calculate the electrical energy consumed in 1 min. in each resistance.​

Answer 1

Answer:

Given

R1 = 3 ohms

R2 = 6 ohms

So, the total resistance = R1 + R2

= 3 ohms + 6 ohms

= 9 ohms

Potential Difference = 12V

Time = 1 min

= 60 seconds

WE KNOW,

P = V^2

R

= 12 X 12

9

= 144

9

= 16

NOW, TO FIND THE ELECTRICAL ENERGY CONSUMED

E = P X t

= 16 X 60

= 960 kWh

Answer 2

Answer :

Two resistors of resistance 3Ω and 6Ω are connected in parallel across a battery of pd 12V.

We have to find electrical energy consumed in 1min in each resistor$.$

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◈ We know that pd remains same across each resistor in parallel connection. Hence pd across both resistors will be same equal to 12V.

❍ Energy consumed by R₁ :

➝ E₁ = V²/R₁ × t

➝ E₁ = 12²/3 × (60)

➝ E₁ = 144/3 × 60

➝ E₁ = 2880J

❍ Energy consumed by R₂ :

➝ E₂ = V²/R₂ × t

➝ E₂ = 12²/6 × (60)

➝ E₂ = 144/6 × 60

➝ E₂ = 1440J