An electric bulb of resistance 400 ohm draws a current of 0.5A.Calculate (a) power of the bulb and potential difference at its end
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Answered by
126
Heya user !!!
Here's the answer you are looking for
The resistance (R) = 400 ohm
and current (I) = 0.5 A
⏩ Power = I²R
= (0.5)²(400)
= 100Watt
⏩ Potential difference (V) = IR
= (0.5)(400)
= 200 V
★★ HOPE THAT HELPS ☺️ ★★
Here's the answer you are looking for
The resistance (R) = 400 ohm
and current (I) = 0.5 A
⏩ Power = I²R
= (0.5)²(400)
= 100Watt
⏩ Potential difference (V) = IR
= (0.5)(400)
= 200 V
★★ HOPE THAT HELPS ☺️ ★★
rapujance:
thanks
welcome
Can you be able to answer this question too pls:Imagine that a space traveller has landed on an unknown planet and he wants to detect the presence of magnetic field on this planet with help of a sensitive galvanometer and a coil of wire.Suggest how he can do that .Give reason for your answer. (b) Name state the phenomenon related to the above situation.(c) Write the name and state the rule that is used to determine the direction of current produced in the above phenomenon
Answered by
34
Power = i²R
= (0.5 A)² × 400 Ω
= 100 Watt
Potential difference = iR
= (0.5 A) × (400 Ω)
= 200 Volts
= (0.5 A)² × 400 Ω
= 100 Watt
Potential difference = iR
= (0.5 A) × (400 Ω)
= 200 Volts
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