Question

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. If the current is supplied by a 12.0 - V battery, what total energy in Joules is delivered to the lightbulb filament during 2.00 s

Answer 1

E = 20.03 J

Step-by-step explanation:

Given that,

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,

Voltage, V = 12 V

We need to find the energy delivered to the lightbulb filament during 2.00 s.

The energy delivered is given by :

`E=I^2Rt`. ....(1)

As,

`I=(q)/(t)\\\\I=(1.67)/(2)\\\\I=0.835\ A`

As per Ohm's law, V = IR

`R=(V)/(I)\\\\R=(12)/(0.835)\\\\R=14.37\ \Omega`

Using formula (1).

`E=0.835^2* 14.37* 2\\\\=20.03\ J`

So, the energy delivered to the lightbulb filament is 20.03 J.