an electric bulb is marked 40 watt and 120 volt it means that in 1 second it converts
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2
P = IV
40= I× 120
I = 40/120
I =0 .33
current, I = Q/t
Q = I × t
0.33 ×1 =Q
Q = 0.33
in 1 second it convert 0.33 coulomb charge into 40 volt power
here is ur answer
I hope it helps you
40= I× 120
I = 40/120
I =0 .33
current, I = Q/t
Q = I × t
0.33 ×1 =Q
Q = 0.33
in 1 second it convert 0.33 coulomb charge into 40 volt power
here is ur answer
I hope it helps you
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Answered by
5
heya ' your answer is
given , power P = 40 watt and volt V = 120 V
therefore ,
current I = P/V = 40/120 =0.33 ampere
now ,
I = Q/t
Q= I x t
in one second
Q = 0.33/1 =0.33 coulomb
hence , in one second it convert 0.33 coulomb charge into 40 watt power
i m right or m i right
given , power P = 40 watt and volt V = 120 V
therefore ,
current I = P/V = 40/120 =0.33 ampere
now ,
I = Q/t
Q= I x t
in one second
Q = 0.33/1 =0.33 coulomb
hence , in one second it convert 0.33 coulomb charge into 40 watt power
i m right or m i right
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