An electric bulb is rated 60W,240V. Calculate its resistance.If the voltage drops to 192V, calculate the power consumed and the current drawn by the bulb. (ASSUMETHATTHERESISTANCE OFTHEBULBREMAINSCHANGED)

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Solution

$P o w e r r a t i n g = 60 w a t t V o l t a g e r a t i n g = 240 v o l t R e s i s \tan c e = \frac{v^{2}}{p} R = \frac{240 \times 240}{60} R = 960 o h m V o l t a g e a c r o s s t h e b u l b = 192 v o l t p o w e r c o n s u m e d = \frac{v o l t a g e^{2}}{R} = \frac{192 \times 192}{960} = 384 w a t t c u r r e n t a c r o s s t h e b u l b = \frac{v o l t a g e}{R} = \frac{192}{960} = 0.2 A$

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