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Alternating Current Applied to an Inductor

A bulb is des...

Question

A bulb is designed to operate at $12 v o l t s$ constant direct current. If this bulb is connected to an alternating current source and gives same brightness. What would be the peak voltage of the source?

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Solution

$E = 12 V (G i v e n) L e t E_{o} b e t h e p e a k v o l t a g e o f t h e b u l b . L e t R a n d T b e t h e r e s i s t a n c e a n d t e m p e r a t u r e o f t h e e l e c t r i c b u l b . S i n c e, h e a t p r o d u c e d i n a r e s i s t o r b y p a s s i n g A l t e r n a t i n g c u r r e n t i s s a m e a s h e a t p r o d u c e d b y d i r e c t c u r r e n t i n a r e s i s t o r i n s a m e t i m e . S o, i^{2} R T = i_{r m s}^{2} R T => \frac{E^{2}}{R^{2}} R T = \frac{E_{r m s}^{2}}{R^{2}} R T => E^{2} = \frac{E_{o}^{2}}{2} (A s w e k n o w, E_{r m s} = \frac{E_{o}}{\sqrt{2}}) => E_{o}^{2} = 2 E^{2} = 2 \times 12^{2} => E_{o} = \sqrt{2 \times 12^{2}} = 12 \sqrt{2} = 16.97 V = 17 V$

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