**Problem 8.28: Find $ v_S(t) $ in the circuit in Fig. P8.28.****Figure P8.28 Description:**This figure represents an electrical circuit composed of:- A voltage source labeled $ v_S(t) $.- A resistor with a resistance of $ 40 \, \Omega $, represented by a zigzag line.- A capacitor with a capacitance of $ 25 \, \mu F $, depicted by two parallel lines with a gap between them, indicating the capacitor's plates.- The voltage across the capacitor is given by the expression $ v_C(t) = 80 \cos (1000t - 60^\circ) \, \text{V} $.The components are connected in series to form a closed loop, with the polarity of the voltage source and the capacitor indicated by plus and minus signs.

Answered: Find vs(t) in the circuit in Fig.

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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**Problem 8.28: Find $ v_S(t) $ in the circuit in Fig. P8.28.**

**Figure P8.28 Description:**

This figure represents an electrical circuit composed of:

- A voltage source labeled $ v_S(t) $.
- A resistor with a resistance of $ 40 \, \Omega $, represented by a zigzag line.
- A capacitor with a capacitance of $ 25 \, \mu F $, depicted by two parallel lines with a gap between them, indicating the capacitor's plates.
- The voltage across the capacitor is given by the expression $ v_C(t) = 80 \cos (1000t - 60^\circ) \, \text{V} $.

The components are connected in series to form a closed loop, with the polarity of the voltage source and the capacitor indicated by plus and minus signs.

Expert Solution

Step 1

Based on given circuit is Series circuit. In series circuits current is same in all elements but voltage is distributed based on their Impedance values.

Here $v_{c} (t)$ is given as

$v_{c} (t) = 80 \cos (1000 t - 60 °) V$

$w = 1000 X_{c} = \frac{1}{j w c} = \frac{1}{j 1000 \times 25 \times 10^{- 6}} = \frac{1000}{j 25} = 40 X_{c} = - j 40$

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