Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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**Problem Statement:** Q1: A current waveform in a 10 µF capacitor is shown below. If the initial voltage in the capacitor is 0.5 mV, find: a) The capacitor voltage and its corresponding waveform. b) The stored energy \( w_c \) in the capacitor at \( t = 3 \) ms. **Diagram Explanation:** The left side of the image depicts a current vs. time graph. The horizontal axis represents time in milliseconds (ms), and the vertical axis represents current in milliamperes (mA). The waveform has the following segments: - From 0 to 1 ms, the current is 10 mA. - From 1 ms to 3 ms, the current is 0 mA. - From 3 ms to 4 ms, the current is -5 mA. **Equations:** 1. The voltage across the capacitor as a function of time: \[ v(t) = \frac{1}{C} \int_{t_0}^{t} i(t) \, dt + V_0 \] 2. The stored energy in the capacitor: \[ w_c = \frac{1}{2} C v^2 \] Where: - \( C \) is the capacitance (10 µF), - \( i(t) \) is the current, - \( V_0 \) is the initial voltage (0.5 mV). **Calculation Steps:** To solve for the voltage and energy, integrate the current over the specified time intervals and apply the formulas provided.
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Transcribed Image Text:**Problem Statement:** Q1: A current waveform in a 10 µF capacitor is shown below. If the initial voltage in the capacitor is 0.5 mV, find: a) The capacitor voltage and its corresponding waveform. b) The stored energy \( w_c \) in the capacitor at \( t = 3 \) ms. **Diagram Explanation:** The left side of the image depicts a current vs. time graph. The horizontal axis represents time in milliseconds (ms), and the vertical axis represents current in milliamperes (mA). The waveform has the following segments: - From 0 to 1 ms, the current is 10 mA. - From 1 ms to 3 ms, the current is 0 mA. - From 3 ms to 4 ms, the current is -5 mA. **Equations:** 1. The voltage across the capacitor as a function of time: \[ v(t) = \frac{1}{C} \int_{t_0}^{t} i(t) \, dt + V_0 \] 2. The stored energy in the capacitor: \[ w_c = \frac{1}{2} C v^2 \] Where: - \( C \) is the capacitance (10 µF), - \( i(t) \) is the current, - \( V_0 \) is the initial voltage (0.5 mV). **Calculation Steps:** To solve for the voltage and energy, integrate the current over the specified time intervals and apply the formulas provided.
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