Question

A 100W electric light bulb is connected to a 200V supply. Calculate (i) the current flowing in the bulb, and (ii) the resistance of the bulb.

Answer 1

Answer:

p=v×i

100=200×i

hence

i=1/2amp

also

p=v(square)/r

r=200x200/100

so

r=400ohm

Answer 2

Given:-

• Power ,P = 100 W
• Potential Difference ,V = 200 v

To Find:-

• (i) the current flowing in the bulb,
• (ii) the resistance of the bulb.

Solution:-

According to the Question

Firstly we calculate the current flowing in the bulb .So using formula

• P = VI

where,

• P denote Power
• V denote Potential Difference
• I denote Current

Substitute the value we get

$:\implies$ 100 = 200 × I

$:\implies$ 200× I = 100

$:\implies$ I = 100/200

$:\implies$ I = 0.5 A

• Hence, the current flowing in the bulb is 0.5 Ampere.

Now, calculating the Resistance of the bulb .

• P = V²/R

Substitute the value we get

$:\implies$ 100 = 200²/R

$:\implies$ 100× R = 200×200

$:\implies$ R = 200×200/100

$:\implies$ R = 200×2

$:\implies$ R = 400 ohm

• Hence, the Resistance of the bulb is 400 ohm.