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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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