**Transcription and Explanation for Educational Website**---### Problem StatementGiven a capacitor of $ C = 7 \, \mu F $, calculate the current going through the capacitor during the interval $ t_1 \leq t \leq t_2 $, when the capacitor voltage is as described in Fig. 6.12. Assume $ V_1 = 6 \, V $, $ V_2 = -9 \, V $, and $ t_1 = 7 \, \mu s $, and $ t_2 = 14 \, \mu s $. **(Note: values might be different than the above problem)**### Graph Description: Figure P6.12The graph in Figure P6.12 illustrates the voltage $ v(t) $ across the capacitor as a function of time $ t $ in microseconds ($\mu s$). Key features of the graph include:- **Voltage Axis (Vertical):** The voltage is represented on the vertical axis in volts (V).- **Time Axis (Horizontal):** The time is indicated on the horizontal axis in microseconds ($\mu s$).**Key Points on the Graph:**1. **At $ t = 0 $:** The voltage starts at the origin.2. **At $ t = t_1 $:** The voltage reaches a peak at $ V1 $, which is equal to 6V.3. **At $ t = t_2 $:** The voltage drops to a trough at $ V2 $, which is equal to -9V.4. **At $ t = t_3 $:** The voltage returns to the level at $ t1 $.**Graph Shape:**- The voltage increases linearly from 0 to $ t_1 $.- It decreases beyond $ t_1 $ to reach $ t_2 $.- There is a linear rise again between $ t_2 $ and $ t_3 $.This graph is essential for understanding how the voltage changes over time, which will help in calculating the current through the capacitor during the specified time interval.

Answered: Given a capacitor of C = 7 µF,…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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**Transcription and Explanation for Educational Website**

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### Problem Statement

Given a capacitor of $ C = 7 \, \mu F $, calculate the current going through the capacitor during the interval $ t_1 \leq t \leq t_2 $, when the capacitor voltage is as described in Fig. 6.12. Assume $ V_1 = 6 \, V $, $ V_2 = -9 \, V $, and $ t_1 = 7 \, \mu s $, and $ t_2 = 14 \, \mu s $. **(Note: values might be different than the above problem)**

### Graph Description: Figure P6.12

The graph in Figure P6.12 illustrates the voltage $ v(t) $ across the capacitor as a function of time $ t $ in microseconds ($\mu s$). Key features of the graph include:

- **Voltage Axis (Vertical):** The voltage is represented on the vertical axis in volts (V).
- **Time Axis (Horizontal):** The time is indicated on the horizontal axis in microseconds ($\mu s$).

**Key Points on the Graph:**

1. **At $ t = 0 $:** The voltage starts at the origin.
2. **At $ t = t_1 $:** The voltage reaches a peak at $ V1 $, which is equal to 6V.
3. **At $ t = t_2 $:** The voltage drops to a trough at $ V2 $, which is equal to -9V.
4. **At $ t = t_3 $:** The voltage returns to the level at $ t1 $.

**Graph Shape:**

- The voltage increases linearly from 0 to $ t_1 $.
- It decreases beyond $ t_1 $ to reach $ t_2 $.
- There is a linear rise again between $ t_2 $ and $ t_3 $.

This graph is essential for understanding how the voltage changes over time, which will help in calculating the current through the capacitor during the specified time interval.

Expert Solution

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We need to determine the current in the capacitor C.

We know

Current in the capacitor C,

$I subscript C equals C fraction numerator d V over denominator d t end fraction space..... left parenthesis 1 right parenthesis w h e r e C equals 7 cross times 10 to the power of negative 6 end exponent F d V equals V subscript 2 minus V subscript 1 d t equals t subscript 2 minus t subscript 1 a n d V subscript 2 equals negative 9 V V subscript 1 equals 6 V t subscript 2 equals 14 cross times 10 to the power of negative 6 end exponent s t subscript 1 equals 7 cross times 10 to the power of negative 6 end exponent s$

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