Question

if the potential difference across a bulb decreases by 2% what will be the change in the power dissipation of the bulb
NEED AN EXPLANATORY ANS
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Answer 1

Answer:

I have given three answer PLEASE check

Explanation:

Well, as you might know power is equal to: P=U⋅I.P=U⋅I.Assuming that the voltage stays constant the power with a drop in current of 2% will be:

P=U⋅I⋅(1−0.02)=U⋅I⋅0.98P=U⋅I⋅(1−0.02)=U⋅I⋅0.98

As you can see, the power will drop by the same amount as the current does, because power is proportional to current.

A voltage drop of 2% on the other hand would cause the power to drop by about 4%, because power is also equal to: P=U2RP=U2R(assuming that the resistance stays constant).

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P=E*I. If the voltage stays the same then the power will be reduced by the same percentage, if loss of current is effected by the increase of resistance due to, heat, however, if the the resistance remains the same then the ohm’s law formula must be used before the power equation. Assuming 120V 60W bulb, the current through the bulb is 2A and the resistance is 60Ω. Now the resistance remains 60Ω and the current is 1.96% (2*0.98), we need to calculate the new voltage. 60Ω*1.96A=177.6V so our new power calculation using P=E*I is P=1.96A*177.6V=230.5W making the change in power not 2% but 3.96%.

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ANSWER

P=I2R%I=2%%P=2(%I)+%R%P=2×2+0=4%

There will be 4% increase in the power of the bulb.