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

Answer:

$\underline{\purple{\ddot{\MasterRohith}}}$

Given:-

• Two bulbs of 100 watt and 200 watt are connected to 220volt battery separately after that both the bulbs are connected in series with a 200 volt battery.

To prove :-

• Find total current flowing in a circuit when both are connected in series .

Explanation :-

• First we should take resistance of 1st bulb .

1. which is

• P1=V^2/R1
• R1=V^2/P1
• R1=(220)^2/100
• R1=484 Ohm.

Now , lets take resistance of 2nd bulb.

• P2=V^2/R2
• R2=V^2/P2
• R2=(220)^2/200
• R2= 242ohm.

Now , total resistance or equivalent resistance for series connection:-

• R=R1+R2
• R=(484+242)ohm.
• R=726ohm.

Therefore ,

By using ohms law ,

• V=IR
• I=V/R
• I=200/726
• I=0.275A.

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Answer 2

Answer:

0.27 ampere is the required answer

Explanation:

According to the Question

It is given that

• Power 1st Blub ,P = 100W
• Power of 2nd bulb ,P' = 200 W
• Potential Difference of 1st bulb ,V = 220v
• Potential Difference of 2nd bulb ,V' = 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 .

• P = V²/R

substitute the value we get

➻ 100 = 220²/R

➻ 100×R = 48400

➻ R = 48400/100

➻ R = 484 ohm

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

Similarly, calculating the resistance of 2nd bulb .

• P' = V'²/R'

Substitute the value we get

➻ 200 = 220²/R'

➻ 200 = 48400/R'

➻ 200×R = 48400

➻ R = 48400/200

➻ R = 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 = R + R'

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

Putting all the value we get

➻ 200 = I × 726

➻ I = 200/726

➻ I = 0.27 A

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