Physics, asked by sunilsihag403, 1 year ago

A bulb is rated 40 watt 220 volt find the current drawn by eat when it is connected to a 220 volt supply also find its resistance if the given blood bulb is replaced by a bulb operating 25 Watt 220 volt will there be any change in the value of current and resistance justify your answer and determine the change

Answers

Answered by abhi178
193

case 1 : power of bulb , P = 40 watt

voltage supplied across the bulb , V = 220V

we know, power supplied by bulb , P = Vi

or, 40 = 220 × i

or, i = 40/220 = 2/11 = 0.18 Amp

again, use formula, P = V²/R to find resistance of bulb

or, P = V²/R

or, R = V²/P

or, R = (220)²/40 = 48400/40 = 1210 ohm

case 2 : if given bulb is replaced by a bulb operating 25 watt , 220 volt.

so, current, I = P'/V = 25/220 = 0.1136 Amp

resistance, R' = V²/P' = (220)²/25 = 1936 ohm.

here it is clearly shown that when we change bulb , current and resistance will be change.

change in current = 0.18 - 0.1136

= 0.0664 Amp [ current decreases when we use 25 watt bulb in place of 40 watt bulb ]

change in resistance = 1936 - 1210= 726 ohm [ resistance increase when we use 25 watt bulb in place of 40 watt ]


sunilsihag403: Thanks bro
Answered by probrainsme101
2

Answer and Explanation:

Case I: Bulb rated as 40 watt 220 volt

Power of bulb, P = 40 watt = 40 W

Potential difference, V = 220 volt = 220 V

As we know, Power = Potential difference × Current

Let the current be I.

Now,

P = V×I

I = P/V

I = 40/220

I = 0.182 ampere = 0.182 A

For finding the resistance of the bulb, we have another formula of power as

P = V²/R

R = V²/P

R = (220)²/40

R = 48400/40

R = 1210 Ω

Case II: Bulb rated as 25 watt 220 volt

Power of bulb, P = 25 watt = 25 W

Potential difference, V = 220 volt = 220 V

As we know, Power = Potential difference × Current

Let the current be I.

Now,

P = V×I

I = P/V

I = 25/220

I = 0.114 ampere = 0.114 A

For finding the resistance of the bulb, we have another formula of power as

P = V²/R

R = V²/P

R = (220)²/25

R = 48400/25

R = 1936 Ω

Difference in current = 0.182 - 0.114 = 0.068 A [Current decreases]

Difference in resistance = 1936 - 1210 = 726 Ω [Resistance increases]

#SPJ3

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