The voltage V across a charged capacitor is given by V(t) = 2e^-0.16t volts, where t is in seconds

(A) what is the voltage V(5) after 5 seconds? Find the time t when the voltage will be 1 volt

Enter your answers to two decimal places

V(5) approximately ___________ volts

t approximately ___________ sec

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

Enter your answer to two decimal places

B approximately ________ %

if t = 5

2 e^-(.16*5)=2 e^-.8 = 2/e^.8 = .899 = .90 volts

1 = 2 e^-(.16t)
ln .5 = -.16 t
t = 4.33 seconds

at 0 sec V = 2
at 5 seconds V = .90

V = Vi (r)^t
.9 = 2 r^5
.45 = r^5
log .45 = 5 log r
r = .852
so every second multiply by .85
so decreases by 15% every second

(A) Let's calculate the voltage V(5) after 5 seconds:

V(t) = 2e^(-0.16t) volts

V(5) = 2e^(-0.16*5)
≈ 2e^(-0.80)
≈ 2*0.4493
≈ 0.8986 volts

Therefore, V(5) is approximately 0.90 volts.

Now, let's find the time t when the voltage will be 1 volt:

V(t) = 2e^(-0.16t) volts

1 = 2e^(-0.16t)

e^(-0.16t) = 1/2

Taking the natural logarithm of both sides:

-0.16t = ln(1/2)

t = ln(1/2) / -0.16
≈ -1.6094 / -0.16
≈ 10.0588 seconds

Therefore, t is approximately 10.06 seconds.

(B) To find the percent decrease each second, we need to calculate the change in voltage per second and express it as a percentage. Let's consider the derivative of V(t):

V'(t) = -0.32e^(-0.16t)

To find the percent decrease, we can divide V'(t) by V(t) and multiply by 100:

B = (V'(t) / V(t)) * 100
= (-0.32e^(-0.16t) / 2e^(-0.16t)) * 100
= -16

Therefore, B is approximately -16%.

(A) To find the voltage V(5) after 5 seconds, plug in t = 5 into the equation V(t) = 2e^(-0.16t):

V(5) = 2e^(-0.16 * 5) volts

Calculating the value:

V(5) = 2 * e^(-0.8)
≈ 2 * 0.4493
≈ 0.8986 volts

So, V(5) is approximately 0.90 volts.

To find the time t when the voltage will be 1 volt, we need to rearrange the equation V(t) = 2e^(-0.16t) and solve for t. Setting V(t) = 1:

1 = 2e^(-0.16t)

Divide both sides by 2:

0.5 = e^(-0.16t)

Take the natural logarithm (ln) of both sides:

ln(0.5) = -0.16t

Solve for t:

t = ln(0.5) / -0.16 ≈ 4.4276 seconds

So, t is approximately 4.43 seconds.

(B) To find the percentage decrease in voltage each second, we can take the derivative of the voltage function V(t) = 2e^(-0.16t) with respect to time:

dV/dt = -0.32e^(-0.16t)

Now, substitute t = 1 into the derivative:

dV/dt at t = 1 = -0.32e^(-0.16 * 1)

Calculating the value:

dV/dt at t = 1 = -0.32e^(-0.16)
≈ -0.32 * 0.8499
≈ -0.27197

To express this decrease as a percentage, we can multiply by 100:

Percent decrease per second = -0.27197 * 100
≈ -27.20%

So, the voltage decreases by approximately 27.20% each second.

To find the voltage V(5) after 5 seconds, we need to substitute t = 5 into the equation V(t) = 2e^-0.16t.

(A)

Step 1: Substitute t = 5 into the equation V(t) = 2e^-0.16t
V(5) = 2e^-0.16(5)

Step 2: Calculate the value of e^-0.16(5) using a calculator.
e^-0.16(5) ≈ 0.3012

Step 3: Multiply 2 by the calculated value.
V(5) ≈ 2 * 0.3012

Step 4: Calculate the value of V(5) approximately.
V(5) ≈ 0.6024 volts

Therefore, V(5) is approximately 0.60 volts.

To find the time t when the voltage will be 1 volt, we need to solve the equation V(t) = 1.

Step 1: Set V(t) = 1 in the equation V(t) = 2e^-0.16t.
1 = 2e^-0.16t

Step 2: Divide both sides of the equation by 2.
e^-0.16t = 0.5

Step 3: Take the natural logarithm of both sides of the equation.
ln(e^-0.16t) = ln(0.5)

Step 4: Use the property of logarithms to simplify the equation.
-0.16t = ln(0.5)

Step 5: Divide both sides of the equation by -0.16.
t = ln(0.5) / -0.16

Step 6: Calculate the value of t approximately using a calculator.
t ≈ 4.3351

Therefore, t is approximately 4.34 seconds.

(B)

The percent decrease B each second can be calculated using the formula:

B = (V(0) - V(1)) / V(0) * 100

Step 1: Calculate the value of V(0) using t = 0 in the equation V(t) = 2e^-0.16t.
V(0) = 2e^-0.16(0)
V(0) = 2e^0
V(0) = 2 * 1
V(0) = 2 volts

Step 2: Calculate the value of V(1) using t = 1 in the equation V(t) = 2e^-0.16t.
V(1) = 2e^-0.16(1)

Step 3: Calculate the value of e^-0.16(1) using a calculator.
e^-0.16(1) ≈ 0.8431

Step 4: Multiply 2 by the calculated value.
V(1) ≈ 2 * 0.8431

Step 5: Calculate the percent decrease B using the formula.
B = (2 - (2 * 0.8431)) / 2 * 100

Step 6: Simplify the equation.
B ≈ (2 - 1.6862) / 2 * 100

Step 7: Calculate the value of B approximately.
B ≈ 15.76%

Therefore, B is approximately 15.76%.