dv + 5v = 10B(t), dt v(0) = 0 %3D For this problem we are just going to look at the equation for the battery. A fully charged battery when connected to the circuit at time zero have a voltage source equation B(t) = 10e-0.5t b. Our battery is off until one second then connects for one second, then is off for one second and does not recharge (it just retains the voltage it had when it was removed from the circuit) and then is turned on for one second, then it turned off for good. Give the function for the battery voltage in the circuit using the Heaviside function and the exponential function.
dv + 5v = 10B(t), dt v(0) = 0 %3D For this problem we are just going to look at the equation for the battery. A fully charged battery when connected to the circuit at time zero have a voltage source equation B(t) = 10e-0.5t b. Our battery is off until one second then connects for one second, then is off for one second and does not recharge (it just retains the voltage it had when it was removed from the circuit) and then is turned on for one second, then it turned off for good. Give the function for the battery voltage in the circuit using the Heaviside function and the exponential function.
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![**Analyzing the Battery Voltage in a Circuit**
For this problem, we are going to examine the equation for the battery.
### Battery Voltage Equation
A fully charged battery, when connected to the circuit at time zero, has a voltage source equation:
\[ B(t) = 10e^{-0.5t} \]
### Initial Conditions
The given differential equation is:
\[ \frac{dv}{dt} + 5v = 10B(t), \]
with the initial condition:
\[ v(0) = 0 \]
### Complex Voltage Function Scenario
**Scenario B:**
Our battery alternates states as follows:
- The battery is off until one second.
- It then connects for one second.
- It is off for one second and does not recharge (retains the voltage it had when removed from the circuit).
- The battery is then turned on for one second.
- Finally, it is turned off for good.
To find the function for the battery voltage in the circuit using the Heaviside function and the exponential function, we need to account for these operational intervals.
The function for the battery voltage, denoted as \( B_2(t) \), can be formulated by segmenting the exponential function according to the specified intervals.
\[ B_2(t) = \begin{cases}
10e^{-0.5(t-1)} & 1 \leq t < 2 \\
0 & \text{otherwise}
\end{cases}
\]
In Heaviside terms (U(t) is the Heaviside step function):
\[ B_2(t) = 10e^{-0.5(t-1)} \left[ U(t-1) - U(t-2) \right] \]
This function uses the Heaviside step function to model the battery switching on and off at the specified times.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd8339843-a6c0-40e6-979c-08fc9d6ca9a2%2Fe236e2f0-6e71-4eac-8342-d59682e3d1ac%2Fx67t7te_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Analyzing the Battery Voltage in a Circuit**
For this problem, we are going to examine the equation for the battery.
### Battery Voltage Equation
A fully charged battery, when connected to the circuit at time zero, has a voltage source equation:
\[ B(t) = 10e^{-0.5t} \]
### Initial Conditions
The given differential equation is:
\[ \frac{dv}{dt} + 5v = 10B(t), \]
with the initial condition:
\[ v(0) = 0 \]
### Complex Voltage Function Scenario
**Scenario B:**
Our battery alternates states as follows:
- The battery is off until one second.
- It then connects for one second.
- It is off for one second and does not recharge (retains the voltage it had when removed from the circuit).
- The battery is then turned on for one second.
- Finally, it is turned off for good.
To find the function for the battery voltage in the circuit using the Heaviside function and the exponential function, we need to account for these operational intervals.
The function for the battery voltage, denoted as \( B_2(t) \), can be formulated by segmenting the exponential function according to the specified intervals.
\[ B_2(t) = \begin{cases}
10e^{-0.5(t-1)} & 1 \leq t < 2 \\
0 & \text{otherwise}
\end{cases}
\]
In Heaviside terms (U(t) is the Heaviside step function):
\[ B_2(t) = 10e^{-0.5(t-1)} \left[ U(t-1) - U(t-2) \right] \]
This function uses the Heaviside step function to model the battery switching on and off at the specified times.
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