Question

Two bulbs of 100watt & 200 watt are connected to 220 volt battery separately , after that both the bulbs are connected in series with a 200 volt battery . find total current flowing in a circuit when both are connected in series .​

Answer 1

Solution :-

It is given that

• Power 1st Blub ,P₁ = 100W
• Power of 2nd bulb ,P₂ = 200 W
• Potential Difference of 1st bulb ,V₁ = 220v
• Potential Difference of 2nd bulb ,V₂ = 200v

we have to calculate the total current flowing in the circuit when both bulbs are connected in series .So , Firstly we calculate the resistance of each bulb .Resistance of 1st bulb

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☄️P₁ = V₁²/R₁

substitute the value we get

➻ 100 = 220²/R₁

➻ 100×R₁ = 48400

➻ R₁ = 48400/100

➻ R₁ = 484 ohm

So, the resistance of 1st bulb is 484 ohms.

Similarly, calculating the resistance of 2nd bulb .

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☄️ P₂ = V₂²/R₂

Substitute the value we get

➻ 200 = 220²/R₂

➻ 200 = 48400/R₂

➻ 200×R₂ = 48400

➻ R₂ = 48400/200

➻ R₂ = 242 ohms

So, the resistance of 2nd bulb is 242 ohms .When these bulb are connected in series . so , the total resistance in series will be

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☄️ R = R₁ + R₂

substitute the value we

➻ R = 242+484

➻ R = 726 ohms

Now , calculating the current flowing through these bulbs . by using ohm's law

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☄️V = IR

Substitute the value we get

➻ 200 = I × 726

➻ I = 200/726

➻ I = 0.27 ampere

• Hence, the current flowing through these bulbs is 0.27 ampere.

Answer 2

Explanation:

Have Given :

→ Power of the 1st bulb = 100 W

→ Power of the 2nd bulb = 200 W

→ Potential difference = 220 V

Resistance of the 1st bulb :

$\rm P_1 = \frac{ {V}^{2} }{ R_1}$

$\rm R_1 = \frac{ {V}^{2} }{ P_1 }$

$\rm R_1 = \frac{{(220)}^{2} }{100}$

$\rm R_1 = \frac{48400 }{100}$

$\rm R_1 = 484 \: Ω$

Resistance of the 2nd bulb :

$\rm P_2 = \frac{V²}{ R_2}$

$\rm R_2 = \frac{V²}{ P_2}$

$\rm R_2= \frac{ (220)²}{ 200}$

$\rm R_2= \frac{ 48400}{ 200}$

$\rm R_2 = 242Ω$

Equivalent resistance for series

connection :

$\rm R = R_1 + R_2$

$\rm R = (484 +242) Ω$

$\rm R = 726 Ω$

Now, we will calculate the current flowing in the circuit for series connection by using 'Ohm's Law'.

$\rm V = IR$

$\rm I = \frac{V}{R}$

$\rm I= \frac{200}{726}$

$\rm I = 0.275 \: A$

I hope it is helpful