What amount of heat will be released in the spiral of an electric lamp in 20 s, if at a voltage of 5 V
| June 27, 2021 | education
What amount of heat will be released in the spiral of an electric lamp in 20 s, if at a voltage of 5 V the current strength in it is 0.2 A?
t = 20 s.
U = 5 V.
I = 0.2 A.
Q -?
The amount of thermal energy Q, which is released in an electric lamp, is expressed by the Joule-Lenz law: Q = I * U * t, where I is the current in the conductor, U is the current voltage at the ends of the conductor, t is the current flow time.
Q = 0.2 A * 5 V * 20 s = 20 J.
Answer: Q = 20 J of heat will be released in the light bulb.
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