Two coils of resistance R1= 3 ohm and R2= 9 ohm are connected in series across a battery of potential difference 14V. Draw the circuit diagram.Find the electrical energy consumed in 1 minute in each resistance?
Answers
Given :
- R₁ = 3 Ω
- R₂ = 9 Ω
- V = 14 V
- Time = 1 minute
To Find :
The electrical energy consumed in 1 minute in each resistance.
Solution :
Analysis :
Here first we have the total resistance. Then by using ohm's law we can find the current as the voltage is given and resistance we are getting in the previous step. Then by using the formula of Joules' Law of heating we can find the electrical energy consumed by each resistance.
Explanation :
Total Resistance :
The two resistors, R₁ and R₂ are in series and the formula is,
Rₛ = R₁ + R₂ +..... Rₙ
where,
- R₁ = 3 Ω
- R₂ = 9 Ω
⇒ Rₛ = (3 + 9) Ω
⇒ Rₛ = 12 Ω
∴ Rₛ = 12 Ω.
The total resistance is 12 Ω.
Now,
According to ohm's law,
V = IR
∴ I = V/R
where,
- V = 14 V
- R = 12 Ω
⇒ I = 14/12
⇒ I = 7/6
∴ I = 7/6 A.
Since the resistance are in series, same current flows in each resistance.
By using Joules' Law of Heating,
H = I²Rt
- 1 minute = 60 sec
Electrical Energy consumed by R₁ (3 Ω) in 60 sec,
⇒ W₁ = I²R₁t
where,
- I = 7/6 A
- R₁ = 3 Ω
- t = 60 sec
⇒ W₁ = (7/6)² × 3 × 60
⇒ W₁ = 49/36 × 3 × 60
⇒ W₁ = 8820/36
⇒ W₁ = 245
∴ W₁ = 245 J.
Electrical Energy consumed by R₂ (9 Ω) in 60 sec,
⇒ W₂ = I²R₂t
where,
- I = 7/6 A
- R₂ = 9 Ω
- t = 60 sec
⇒ W₂ = (7/6)² × 9 × 60
⇒ W₂ = 49/36 × 9 × 60
⇒ W₂ = 26460/36
⇒ W₂ = 735
∴ W₂ = 735 J.
Electrical energy consumed by 3 Ω resistor is 245 J.
Electrical energy consumed by 9 Ω resistor is 735 J.
