Physics, asked by jismaria19, 9 hours ago

an electric bulb was designed to deliver certain power at 250 v . when this bulb was used to 100 v , the power it delivered was 10w. what was the power of which it was designed initially

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Answers

Answered by NewGeneEinstein
3

Answer:-

  • Power=P1=?
  • Power=P2=10watt
  • Voltage=V1=250V
  • Voltage=V2=100V

Here Current (I) is constant

We know

\boxed{\sf P=VI}

\\ \sf\longmapsto I=\dfrac{P}{V}

  • Comparing both

\\ \sf\longmapsto \dfrac{P_1}{V_1}=\dfrac{P_2}{V_2}

\\ \sf\longmapsto \dfrac{P_1}{250}=\dfrac{10}{100}

\\ \sf\longmapsto P_1=\dfrac{250\times 10}{100}

\\ \sf\longmapsto P_1=\dfrac{2500}{100}

\\ \sf\longmapsto P_1=25W

\\ \underline{\boxed{\bf{\therefore P_1=25W}}}

Answered by maryamasimughal48
1

Given:

Voltage1=250V

Voltage2=100V

Power2=10W

To Find:

Power1= ?

Calculations:

As we know

P=IV

So,

I=P/V

I=P1/V1. (I)

I=P2/V2. (II)

Comparing both equations:

P1/V1=P2/V2

P1= P2×V1/V2

P1= 10W×250V/100V

P1= 2500/100

P1 = 25W

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