- The effective duration of the flash is $0.21 s$, which we can assume is the time constant of the capacitor, during which it produces an average of $35 mW$ from an average voltage of $2.85 V$.
How many coulombs of charge move through the light?
In this question, we have to find the charge in coulombs during the flash of a given light having a voltage of $2.85 V$
We should recall that a current is the flow rate of electrons in the conductor and its SI unit is $Ampere$, represented by the letter $A$.
Expert Answer
The electric current applied across linear resistance is directly proportional to the voltage applied across it at a constant temperature. This is known as Ohm’s Law, and it is represented as:
\[V = I \times R\]
To find the charge $Q$, we have the following formula:
\[I = Q/t\]
writing in terms of $Q$:
\[Q= I \times t\]
Here,
$Q$ is the required charge in coulombs
$I$ is the current in amperes
$t$ is the time in sec
As we don’t have the value of current $I$ given in the question but we know that current is equal to power divided by voltage, that is:
\[I = P/V\]
Here
$I$ is current
$P$ is power in watts
and $V$ is voltage
Putting in the above equation, we get:
\[Q = (P/V) \times t\]
Substituting the values in the above equation:
\[Q = {\frac{3.5 \times 10^{-1}}{2.85}} \times 0.21 \]
\[Q = 5.8510 \times 10^{-1} C\]
Numerical Answer
So the value of the charge that moves through the light during $0.21 s$ of flash comes out to be
\[Q = 5.8510 \times 10^{-1} C\].
Example
The effective duration of the flash is $0.25 s$, which we can assume is the time constant of the capacitor, during which it produces an average of $65 mW$ from an average voltage of $2.85 V$.
How much energy in joules does it dissipate? Also, find the coulombs of charge that move through the light.
Given as:
$t = 0.25 s $
$P= 65 \times 10^{-3} W$
$V=2.85 V$
To calculate energy, we have the formula as follows:
\[E = P \times t \]
Putting values in the equation above, we get:
\[E = 0.01625 J \]
To calculate charge $Q$, we have:
\[Q = E/V \]
\[Q = 0.01625 \]
\[P = \frac {0.01625}{2.85} \]
The value of the charge that moves through the light during $0.25 s$ of flash comes out to be
\[Q = 5.701 \times 10^{-3} C \].