To find the mean current in the given setup, we need to follow a series of steps using the relevant physics equations. Let's break it down:
Firstly, we are given:
- Potential difference (V) = 200 volts (V)
- Time (t) = [tex]\(2.8 \times 10^{-4}\)[/tex] seconds (s)
- Energy (E) = 1.4 joules (J)
1. Finding the charge moved, [tex]\(Q\)[/tex]:
The relationship between energy (E), potential difference (V), and charge (Q) is given by the equation:
[tex]\[
E = V \times Q
\][/tex]
Rearrange to solve for [tex]\(Q\)[/tex]:
[tex]\[
Q = \frac{E}{V}
\][/tex]
Plugging in the given values:
[tex]\[
Q = \frac{1.4 \text{ J}}{200 \text{ V}}
\][/tex]
[tex]\[
Q = 0.007 \text{ coulombs (C)}
\][/tex]
2. Finding the mean current, [tex]\(I\)[/tex]:
Current (I) is the rate of flow of charge and is defined by the equation:
[tex]\[
I = \frac{Q}{t}
\][/tex]
Using the charge [tex]\(Q\)[/tex] we found earlier and the given time [tex]\(t\)[/tex]:
[tex]\[
I = \frac{0.007 \text{ C}}{2.8 \times 10^{-4} \text{ s}}
\][/tex]
[tex]\[
I = 25 \text{ amperes (A)}
\][/tex]
Therefore, the mean current when the flash unit is used is [tex]\(25 \text{ A}\)[/tex].
