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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.

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

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

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