A torch bulb is rated as 2.5 V, 500mA.Find its (i) Power (ii)Resistance

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Solution

Dear Student ,
$(i) P o w e r = P = V I = 2 \cdot 5 \times 500 \times 10^{- 3} = 1 \cdot 25 W (i i) R e s i s \tan c e = R = \frac{P}{I^{2}} = 5 Ω$
Regards

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A torch bulb is rated as 2.5 V, 500mA.Find its (i) Power (ii)Resistance

Dear Student , iPower=P=VI=2·5×500×10-3=1·25 WiiResistance=R=PI2=5 Ω Regards

Dear Student ,
$(i) P o w e r = P = V I = 2 \cdot 5 \times 500 \times 10^{- 3} = 1 \cdot 25 W (i i) R e s i s \tan c e = R = \frac{P}{I^{2}} = 5 Ω$
Regards