We are given the equation for an alternating voltage:

$v(t) = 50 \sin(40t)$

where $v(t)$ is the voltage in volts, and $t$ is the time in seconds.

To find the rate of change of voltage at $t = 20 \times 10^{-3}$ seconds:

The rate of change of voltage is the derivative of the voltage function with respect to time. So, we first differentiate $v(t)$ with respect to $t$:

$\frac{dv}{dt} = \frac{d}{dt} \left( 50 \sin(40t) \right)$

Using the chain rule of differentiation:

$\frac{dv}{dt} = 50 \cdot \cos(40t) \cdot 40 = 2000 \cos(40t)$

Now, we need to evaluate this at $t = 20 \times 10^{-3}$ seconds, or $t = 0.02$ seconds.

Substituting $t = 0.02$:

$\frac{dv}{dt} = 2000 \cos(40 \times 0.02)$

$\frac{dv}{dt} = 2000 \cos(0.8)$

Using a calculator:

$\cos(0.8) \approx 0.6967$

Now:

$\frac{dv}{dt} = 2000 \times 0.6967 = 1393.4 \, \text{V/s}$

Thus, the rate of change of voltage at $t = 0.02$ seconds is approximately 1393.4 V/s.

Answer:

The correct choice is 1393.4 V/s.

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Would you like further details or clarification? Here are 5 related questions:

1. How do you calculate the rate of change of a sinusoidal function?
2. What would the voltage be at $t = 0.02$ seconds?
3. How does the rate of change of voltage vary over time for sinusoidal signals?
4. How would you find the maximum rate of change for this alternating voltage?
5. What other types of alternating voltage signals could you work with, and how would you differentiate them?

Tip: When differentiating trigonometric functions involving time, always apply the chain rule carefully to account for any constants within the argument of the sine or cosine function.