Question

An electric power source supplies a constant current of 4 A for 6.5 s to a light bulb. The total energy given off in the form of light and heat is 2.99 kJ. Calculate the voltage drop across the bulb.

Answer 1

Answer:

given : P=2.99 kj , I=4 A , t=6.5s

P= W/ t

P= q∆v/t

P=It∆v/t

P=I∆v

∆v= 2.99*10³/4=0.7475*10³ V

please! mark it brainlist...