Question

The voltage V across a charged capacitor after t seconds is given by V(t) = 7e^{-0.5t}. By what percentage does the voltage decrease each second? (Round your answer to the nearest thousandth.)

Answer 1

The voltage decreases by approximately 47.46% each second.

Step-by-step explanation:

The voltage V across a charged capacitor after t seconds is given by V(t) = 7e-0.5t. To find the percentage by which the voltage decreases each second, we can calculate the difference in voltage after one second. Let's calculate:

V(1) = 7e-0.5(1) = 7e-0.5 = 3.678

The voltage decreases from 7 to 3.678, which is a decrease of approximately 3.322 units. To find the percentage decrease, we divide this value by the initial voltage and then multiply by 100:

Percentage decrease = (3.322/7) * 100 = 47.46%

Therefore, the voltage decreases by approximately 47.46% each second.