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A 500W electric heater is designed to work with a 115V line. If the voltage of line drops to 110V, then what will be percentage loss of energy?
Can u pls guide through?

Rakshanda Valanju , 6 Years ago
Grade 12
anser 1 Answers
Arun

Last Activity: 6 Years ago

Dear student
 

Power dissipiated in external resistance -

P=(\frac{E}{R+r})^{2}R

P_{i}=\frac{v_{i}^{2}}{R}= \frac{\left ( 115 \right )^{2}}{R} - - -- - -- - (1)

P_{f}=\frac{v_{f}^{2}}{R}= \frac{\left ( 110 \right )^{2}}{R}

Change =\frac{P_{i}-P_{f}}{P_{i}}= 1-\frac{P_{f}}{P_{i}}= 1-\left ( \frac{110}{115} \right )^{2}

=\frac{5\times 225}{115 \times 115}=8.5\%

 
Regards
Arun (askIITians forum expert)

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