The mains voltage in British household varies with time that v = 339.5sin(100π t) where t is the time from the start of a cycle. What is the voltage when t = 0.005* seconds?

The given equation for voltage is:

\[
v(t) = 339.5 \sin(100\pi t)
\]

where \( t \) is the time in seconds, and the amplitude is 339.5 V. 

To find the voltage when \( t = 0.005 \) seconds, we substitute \( t = 0.005 \) into the equation:

\[
v(0.005) = 339.5 \sin(100\pi \times 0.005)
\]

Now let's calculate the result.

\[
v(0.005) = 339.5 \sin(0.5\pi)
\]

Since \( \sin(0.5\pi) = 1 \), the voltage is:

\[
v(0.005) = 339.5 \times 1 = 339.5 \, \text{V}
\]

So, the voltage at \( t = 0.005 \) seconds is **339.5 V**.

Would you like more details or have any other questions?

Here are 5 related questions to consider:
1. What is the frequency of the AC voltage in this system?
2. How would you find the time when the voltage first reaches its maximum value?
3. What is the period of the sine wave represented by this voltage equation?
4. How can you calculate the root mean square (RMS) voltage for this mains voltage?
5. What would the voltage be at \( t = 0.002 \) seconds?

**Tip**: The sine function is periodic, so knowing its period helps in predicting voltage values at various times!

The mains voltage in British household varies with time that v = 339.5sin(100π t) where t is the time from the start of a cycle. What is the voltage when t = 0.005 seconds?

Explore how to calculate the voltage in a British household's AC mains system when the voltage varies with time according to the equation v = 339.5sin(100π t). Learn how to apply trigonometric functions to determine the voltage at t = 0.005 seconds, and understand related concepts such as sine waves and periodic functions.

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