5. What would the voltage be at the following times if the initial voltage is 8V, the resistance is 2 kf2 and the capacitance is 50 µF. (a) At t=0 ms, V (b) At t-100 ms, V= (c) At t-1000 ms, V=
5. What would the voltage be at the following times if the initial voltage is 8V, the resistance is 2 kf2 and the capacitance is 50 µF. (a) At t=0 ms, V (b) At t-100 ms, V= (c) At t-1000 ms, V=
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![**Question 5: Voltage Calculation Over Time in an RC Circuit**
What would the voltage be at the following times if the initial voltage is 8V, the resistance is 2 kΩ, and the capacitance is 50 μF?
(a) At t = 0 ms, V = ______________
(b) At t = 100 ms, V = ______________
(c) At t = 1000 ms, V = ______________
---
*Note: To calculate the voltage across the capacitor at a given time in an RC circuit, use the formula:*
\[ V(t) = V_0 \cdot e^{-\frac{t}{RC}} \]
*where \( V_0 \) is the initial voltage, \( t \) is the time, \( R \) is the resistance, and \( C \) is the capacitance.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93d9f4fa-0622-4417-979d-12a19703a696%2F43d1c0c0-65e3-4cdf-99f3-18389176ea33%2Ffbrdp1_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 5: Voltage Calculation Over Time in an RC Circuit**
What would the voltage be at the following times if the initial voltage is 8V, the resistance is 2 kΩ, and the capacitance is 50 μF?
(a) At t = 0 ms, V = ______________
(b) At t = 100 ms, V = ______________
(c) At t = 1000 ms, V = ______________
---
*Note: To calculate the voltage across the capacitor at a given time in an RC circuit, use the formula:*
\[ V(t) = V_0 \cdot e^{-\frac{t}{RC}} \]
*where \( V_0 \) is the initial voltage, \( t \) is the time, \( R \) is the resistance, and \( C \) is the capacitance.*
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![### Problem Statement:
What would the voltage be at the following times if the initial voltage is 8V, the resistance is 2 kΩ, and the capacitance is 50 μF?
### Solution:
**(a) At t=0 ms, V =**
The initial voltage at time t=0 ms is given by the formula:
\[ v = V(1 - e^{-\frac{t}{RC}}) \]
Using the initial conditions:
\[ V = 8V, \]
\[ t = 0 ms, \]
\[ R = 2000 \, \Omega, \]
\[ C = 50 \, \mu F. \]
Substitute into the equation:
\[ v = 8(1 - e^{0}) = 0V. \]
Corrected value:
\[ V = 8V. \]
**(b) At t=100 ms, V =**
Using the same formula:
\[ v = 8(1 - e^{-\frac{100 \times 10^{-3}}{2000 \times 50 \times 10^{-6}}}) \]
\[ v = 8(1 - e^{-\frac{100 \times 10^{-3}}{100}}) \]
\[ v = 8(1 - e^{-0.1}) \]
\[ v \approx 5.057V. \]
**(c) At t=1000 ms, V =**
Using the formula again:
\[ v = 8(1 - e^{-\frac{1000 \times 10^{-3}}{2000 \times 50 \times 10^{-6}}}) \]
\[ v = 8(1 - e^{-\frac{1000 \times 10^{-3}}{100}}) \]
\[ v = 8(1 - e^{-1}) \]
\[ v \approx 8V. \]
There is a question mark next to this final calculation, indicating uncertainty or a need for verification. The expected value should be checked for correctness based on the provided formula and assumptions.](https://content.bartleby.com/qna-images/question/93d9f4fa-0622-4417-979d-12a19703a696/6a5e0ced-0567-493f-85f6-7af2c0071bbf/mv8jsir_processed.png)
Transcribed Image Text:### Problem Statement:
What would the voltage be at the following times if the initial voltage is 8V, the resistance is 2 kΩ, and the capacitance is 50 μF?
### Solution:
**(a) At t=0 ms, V =**
The initial voltage at time t=0 ms is given by the formula:
\[ v = V(1 - e^{-\frac{t}{RC}}) \]
Using the initial conditions:
\[ V = 8V, \]
\[ t = 0 ms, \]
\[ R = 2000 \, \Omega, \]
\[ C = 50 \, \mu F. \]
Substitute into the equation:
\[ v = 8(1 - e^{0}) = 0V. \]
Corrected value:
\[ V = 8V. \]
**(b) At t=100 ms, V =**
Using the same formula:
\[ v = 8(1 - e^{-\frac{100 \times 10^{-3}}{2000 \times 50 \times 10^{-6}}}) \]
\[ v = 8(1 - e^{-\frac{100 \times 10^{-3}}{100}}) \]
\[ v = 8(1 - e^{-0.1}) \]
\[ v \approx 5.057V. \]
**(c) At t=1000 ms, V =**
Using the formula again:
\[ v = 8(1 - e^{-\frac{1000 \times 10^{-3}}{2000 \times 50 \times 10^{-6}}}) \]
\[ v = 8(1 - e^{-\frac{1000 \times 10^{-3}}{100}}) \]
\[ v = 8(1 - e^{-1}) \]
\[ v \approx 8V. \]
There is a question mark next to this final calculation, indicating uncertainty or a need for verification. The expected value should be checked for correctness based on the provided formula and assumptions.
Solution
by Bartleby ExpertFollow-up Question
Transcribed Image Text:5. What would the voltage be at the following times if the initial voltage is 8V, the resistance is 2 k and
the capacitance is 50 uF.
You
At t=0 ms, V =
Ov 8V
(b) At t=100 ms, V=_S.SV
At t=1000 ms, V=
8V
V = Vu (1-22)-> (8)(1-eft) = uv
-100
v = (8) 11-e (2000 (50x10²4)) -- S.USTV
-1000103 +0₁) = 8.00
V= (8) (1-e 2000 (Saxiy
2
0
Solution
by Bartleby ExpertFollow-up Question
![**Problem 5: Voltage Calculation with Initial Conditions**
Given:
- Initial voltage: 8V
- Resistance: 2 kΩ
- Capacitance: 50 μF
**Objective:**
Determine the voltage at specific time intervals.
### (a) At \( t = 0 \) ms, \( V = \)
\[ V = V_0(1 - e^{-\frac{t}{RC}}) \]
\[ V = 8(1 - e^{-\frac{0}{2000 \times 50 \times 10^{-6}}}) = 0 \text{ V} \]
### (b) At \( t = 100 \) ms, \( V = \)
\[ V = 8(1 - e^{-\frac{100 \times 10^{-3}}{2000 \times 50 \times 10^{-6}}}) \approx 5.05 \text{ V} \]
### (c) At \( t = 1000 \) ms, \( V = \)
\[ V = 8(1 - e^{-\frac{1000 \times 10^{-3}}{2000 \times 50 \times 10^{-6}}}) = 8.00 \text{ V} \]
\S, the voltage at each time interval is calculated using the charging formula for a capacitor in an RC circuit, where \( V_0 \) is the initial voltage, \( R \) is the resistance, and \( C \) is the capacitance.](https://content.bartleby.com/qna-images/question/93d9f4fa-0622-4417-979d-12a19703a696/6e314eec-9bf2-415d-b5b2-5c395e0e3e2b/nehr10e_processed.png)
Transcribed Image Text:**Problem 5: Voltage Calculation with Initial Conditions**
Given:
- Initial voltage: 8V
- Resistance: 2 kΩ
- Capacitance: 50 μF
**Objective:**
Determine the voltage at specific time intervals.
### (a) At \( t = 0 \) ms, \( V = \)
\[ V = V_0(1 - e^{-\frac{t}{RC}}) \]
\[ V = 8(1 - e^{-\frac{0}{2000 \times 50 \times 10^{-6}}}) = 0 \text{ V} \]
### (b) At \( t = 100 \) ms, \( V = \)
\[ V = 8(1 - e^{-\frac{100 \times 10^{-3}}{2000 \times 50 \times 10^{-6}}}) \approx 5.05 \text{ V} \]
### (c) At \( t = 1000 \) ms, \( V = \)
\[ V = 8(1 - e^{-\frac{1000 \times 10^{-3}}{2000 \times 50 \times 10^{-6}}}) = 8.00 \text{ V} \]
\S, the voltage at each time interval is calculated using the charging formula for a capacitor in an RC circuit, where \( V_0 \) is the initial voltage, \( R \) is the resistance, and \( C \) is the capacitance.
Solution
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