The voltage V across a charged capacitor is given by V(t)6e(-.26)t volts, where t is in seconds.

Question

The voltage V across a charged capacitor is given by V(t)6e^(-.26)t volts, where t is in seconds.

 

(a) What is the voltage V(6) after 6 seconds? Find the time t when the voltage will be 1 volt.

 

Received 1.26 by Googling, but not sure about how to achieve this or any of it.

Round your answers to two decimal places.

 

(b) By what percent does the voltage decrease each second?

Round your answer to two decimal places.

 

 

Answer 1

I will correct you a little on this. You seem to be missing an equal sign in your function.

 

V(t) = 6e^-.26t, t in seconds

 

(a)  V(6) = 6e^-.26(6) ≈ 1.26 volts

 

      The time when the voltage will be 1 volt is found by solving V(t) = 1 for t.

 

      1 = 6e^-.26t ⇒ e^-.26t = 1/6 ⇒ -.26t = ln (1/6) = -ln 6 ⇒ t = (ln 6)/.26 = 6.89 seconds.

 

(b)  The key to this is to locate the coefficient of t in the exponent for V(t). This gives the rate at which the voltage changes as a decimal. Convert it to a percent, and you have that the voltage decreases by 26% each second.

 

You can check this by computing the voltages at 1 second time intervals and computing the percent difference.

 

V(0) = 6

 

V(1) = 4.63 and the percent difference is (4.63 - 6)/6 = -.23 or decreasing at 23%

 

V(2) = 3.57 and the percent difference is (3.57 - 4.63)/4.63 = -.23 or decreasing at 23%

 

V(3) = 2.75 and the percent difference is (2.75 - 3.57)/3.57 = -.23 or decreasing at 23%

 

V(4) = 2.12 gives a percent difference of 23%

 

These are only approximate percent differences. the exact voltage rate of decrease must be found using Calculus, and I assume you haven't learned that yet.