Question

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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Answer 1

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

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

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

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

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

$\\ \sf\longmapsto P_1=25W$

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

Answer 2

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