Question

Two coils of resistance 3 ohm and 6 ohm are connected in series across a battery of 12 volt find the electrical energy consumed in one minute in each resistance when these are connected in series ??
Answer it​

Answer 1

Answer:

Given conditions ⇒

R₁ = 3 Ω

R₂ = 6 Ω

A/c to the Question, resistors are connected in series,

∴ Req. = R₁ + R₂

 = 3 + 6

 = 9 Ω

Using the Ohm's law,

V = I × R

∴ I = V/R

∴ I = 12/9  [∵ V = 12 V]

∴ I = 4/3 A.

We know, the Current remains same in the series, i.e., If the Resistors are connected in the series, then the same current will flow through all the resistors.

∴ In Case of R₁

R₁ = 3 Ω

I = 4/3 Ω

∴ Potential in R₁ = 4/3 × 3 [ From Ohm's law]

= 4 V

Now, Power = V₁ × I

= 4 × 4/3

= 16/3 W.

∴ Electrical Energy Consumed = Power × time

 = 16/3 × 60

 = 16 × 20

 = 320 J.

Hence, the Electrical Energy consumed in R₁ is 320 J.

In Case of R₂,

R₂ = 6 Ω

I = 4/3

∴ V₂ = 6 × 4/3

= 8 V.

∴ Power = 8 × 4/3

  = 32/3 W.

∴ Electrical Energy Consumed = 32/3 × 60

 = 32 × 20

 = 640 J.

Hence, the Electrical Energy consumed in each resistors is 640 J.

Hope it helps.

Answer 2

Given conditions ⇒

R₁ = 3 Ω

R₂ = 6 Ω

A/c to the Question, resistors are connected in series,

∴ Req. = R₁ + R₂

= 3 + 6

= 9 Ω

Using the Ohm's law,

V = I × R

∴ I = V/R

∴ I = 12/9 [∵ V = 12 V]

∴ I = 4/3 A.

We know, the Current remains same in the series, i.e., If the Resistors are connected in the series, then the same current will flow through all the resistors.

∴ In Case of R₁

R₁ = 3 Ω

I = 4/3 Ω

∴ Potential in R₁ = 4/3 × 3 [ From Ohm's law]

= 4 V

Now, Power = V₁ × I

= 4 × 4/3

= 16/3 W.

∴ Electrical Energy Consumed = Power × time

= 16/3 × 60

= 16 × 20

= 320 J.

Hence, the Electrical Energy consumed in R₁ is 320 J.

In Case of R₂,

R₂ = 6 Ω

I = 4/3

∴ V₂ = 6 × 4/3

= 8 V.

∴ Power = 8 × 4/3

= 32/3 W.

∴ Electrical Energy Consumed = 32/3 × 60

= 32 × 20

= 640 J.

Hence, the Electrical Energy consumed in each resistors is 640 J.

..