# The strength of the current in the electric lamp of the floodlight is 2 A. How much voltage is supplied

The strength of the current in the electric lamp of the floodlight is 2 A. How much voltage is supplied to the floodlight if it consumes 45.6 kJ in 1 minute.

Initial data: I (current in the lamp of the searchlight) = 2 A; A (the energy consumed by the searchlight during the time t) = 45.6 kJ = 45.6 * 10 ^ 3 J; t (floodlight operating time) = 1 min = 60 s.

The power of an electric lamp of a spotlight can be calculated using the formulas:

P = A / t and P = U * I, whence A / t = U * I and U = A / (t * I).

Let’s perform the calculation:

U = 45.6 * 10 ^ 3 / (60 * 2) = 380 V.

Answer: The voltage on the light bulb of the spotlight is 380 V.

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