Question

2 coils of resistance 3 ohm and 6 ohm are connected in series with a battery of voltage 12 volts. Find the electrical energy consumed in 1 minute in each resistance when these are connected in series.

Answer 1

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

Answer:

hope this answer will help