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Figure 17-9 C = 100 µuF %3D V = 12 V 12-V lamp S, 17-2 In Fig. 17–9, explain why the bulb will light for just an instant when S, is initially closed.

Figure 17-9 C = 100 µuF %3D V = 12 V 12-V lamp S, 17-2 In Fig. 17–9, explain why the bulb will light for just an instant when S, is initially closed.

Introductory Circuit Analysis (13th Edition)
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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**Figure 17-9** This figure illustrates a simple electrical circuit comprising a 12-volt battery (\( V_T = 12 \, \text{V} \)), a 100 microfarad capacitor (\( C_1 = 100 \, \mu\text{F} \)), a switch (\( S_1 \)), and a 12-volt lamp. - **Battery (\( V_T \))**: Provides a 12-volt potential difference that powers the circuit. - **Capacitor (\( C_1 \))**: Stores electrical energy, with a capacitance of 100 microfarads. - **Switch (\( S_1 \))**: Used to open or close the circuit. Initially open, it completes the circuit when closed, allowing current to flow. - **12-V Lamp**: Lights up when sufficient current passes through it. --- **Exercise 17-2** Explain why the bulb will light for just an instant when \( S_1 \) is initially closed. This scenario involves a transient response in the circuit. When \( S_1 \) is closed, the capacitor \( C_1 \) begins to charge. Initially, it acts like a short circuit, allowing current to flow and light the bulb. As the capacitor charges, the current decreases until it stops, at which point the bulb goes out.
Transcribed Image Text:**Figure 17-9** This figure illustrates a simple electrical circuit comprising a 12-volt battery (\( V_T = 12 \, \text{V} \)), a 100 microfarad capacitor (\( C_1 = 100 \, \mu\text{F} \)), a switch (\( S_1 \)), and a 12-volt lamp. - **Battery (\( V_T \))**: Provides a 12-volt potential difference that powers the circuit. - **Capacitor (\( C_1 \))**: Stores electrical energy, with a capacitance of 100 microfarads. - **Switch (\( S_1 \))**: Used to open or close the circuit. Initially open, it completes the circuit when closed, allowing current to flow. - **12-V Lamp**: Lights up when sufficient current passes through it. --- **Exercise 17-2** Explain why the bulb will light for just an instant when \( S_1 \) is initially closed. This scenario involves a transient response in the circuit. When \( S_1 \) is closed, the capacitor \( C_1 \) begins to charge. Initially, it acts like a short circuit, allowing current to flow and light the bulb. As the capacitor charges, the current decreases until it stops, at which point the bulb goes out.
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