Answer:
E = 20.03 J
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 :
[tex]E=I^2Rt[/tex]. ....(1)
As,
[tex]I=\dfrac{q}{t}\\\\I=\dfrac{1.67}{2}\\\\I=0.835\ A[/tex]
As per Ohm's law, V = IR
[tex]R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.835}\\\\R=14.37\ \Omega[/tex]
Using formula (1).
[tex]E=0.835^2\times 14.37\times 2\\\\=20.03\ J[/tex]
So, the energy delivered to the lightbulb filament is 20.03 J.
