Question

If the initial voltage is 12 kV, how long does it take to decline to 6.00 × 10^2 V in milliseconds?

Answer 1

To find the time it takes for the voltage to decline from 12 kV to 6.00 × 10^2 V, we can use the concept of the time constant. The time constant (t) is given by the equation t = RC, where R is the resistance and C is the capacitance. From the given information, we have R = 1.00 × 10³ ? and t = 8.00 ms (as calculated before). So, we can solve for C as C = t/R. Plugging in the values, we get C = 8.00 ms / 1.00 × 10³ ? = 8.00 × 10⁻⁶ F. Now, we have the time constant (t) and the initial voltage (V₀) and want to find the time (t') it takes for the voltage to decline to 6.00 × 10^2 V. The equation for the voltage as a function of time is given by V = V₀e^(-t/t'), where V₀ is the initial voltage, t is the time, and t' is the time constant. Rearranging the equation to solve for t', we get t' = -t / ln(V/V₀). Plugging in the values, we have t' = -8.00 ms / ln((6.00 × 10^2 V) / (12 kV)) = 5.531 ms. Therefore, it takes approximately 5.531 milliseconds for the voltage to decline from 12 kV to 6.00 × 10^2 V.

Step-by-step explanation:

To find the time it takes for the voltage to decline from 12 kV to 6.00 × 10^2 V, we can use the concept of the time constant. The time constant (t) is given by the equation t = RC, where R is the resistance and C is the capacitance. From the given information, we have R = 1.00 × 10³ ? and t = 8.00 ms (as calculated before). So, we can solve for C as C = t/R. Plugging in the values, we get C = 8.00 ms / 1.00 × 10³ ? = 8.00 × 10⁻⁶ F.

Now, we have the time constant (t) and the initial voltage (V₀) and want to find the time (t') it takes for the voltage to decline to 6.00 × 10^2 V. The equation for the voltage as a function of time is given by V = V₀e^(-t/t'), where V₀ is the initial voltage, t is the time, and t' is the time constant. Rearranging the equation to solve for t', we get t' = -t / ln(V/V₀). Plugging in the values, we have t' = -8.00 ms / ln((6.00 × 10^2 V) / (12 kV)) = 5.531 ms. Therefore, it takes approximately 5.531 milliseconds for the voltage to decline from 12 kV to 6.00 × 10^2 V.