# A Christmas light is made to flash via the discharge of a capacitor

• 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 $A$.

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:

$Q = {\frac{3.5 \times 10^{-1}}{2.85}} \times 0.21$

$Q = 5.8510 \times 10^{-1} C$

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

$Q = 5.701 \times 10^{-3} C$.