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

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 :

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

As,

$I=\dfrac{q}{t}\\\\I=\dfrac{1.67}{2}\\\\I=0.835\ A$

As per Ohm's law, V = IR

$R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.835}\\\\R=14.37\ \Omega$

Using formula (1).

$E=0.835^2\times 14.37\times 2\\\\=20.03\ J$

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