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...
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Answer 1

Answer:

The expression for the equivalent resistance in series combination is as follows;

R = R1+R2

Here, and are the resistances.

Put and .R1 = 3 Ohm and R2 = 6ohm

R = 3 Ohm + 6 ohm

R = 9ohm

The expression for the power in terms of voltage and resistance is as follows;

P = V2/R

Put V= 12 V  and R= 9 ohm.

P = (12)^2/9

P= 16 W

The expression for the electrical energy in terms of power and time is as follows;

E=Pt

Here, P is the power and t is the time.

Convert the given time into hour by dividing time 1 min by 60.

Put P= 16 W and t =1 min.

E=(16)(1\60)

E=(16)(1\60)

E= 960 kWh

Therefore, the electrical energy is 960 kWh.

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.