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2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K, = 2 and K, = 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The distance is d= 0.04 m. A potential difference of AV = 24 V is applied to the circuit. |(a) Find the equivalent capacitance of the system. (b) Find the potential AV,. (E=8.85×10-12 C²/N×m²) A/2 AV1 K2 A/2 A d K2 KI K1 A/2 2d/3 d/3 K2 A/2 d. AV = 24 V -------

2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K, = 2 and K, = 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The distance is d= 0.04 m. A potential difference of AV = 24 V is applied to the circuit. |(a) Find the equivalent capacitance of the system. (b) Find the potential AV,. (E=8.85×10-12 C²/N×m²) A/2 AV1 K2 A/2 A d K2 KI K1 A/2 2d/3 d/3 K2 A/2 d. AV = 24 V -------

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K, = 2 and K, = 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The distance is d= 0.04 m. A potential difference of AV = 24 V is applied to the circuit. |(a) Find the equivalent capacitance of the system. (b) Find the potential AV,. (E=8.85×10-12 C²/N×m²) A/2 AV1 K2 A/2 A d K2 KI K1 A/2 2d/3 d/3 K2 A/2 d. AV = 24 V -------
Transcribed Image Text:2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K, = 2 and K, = 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The distance is d= 0.04 m. A potential difference of AV = 24 V is applied to the circuit. |(a) Find the equivalent capacitance of the system. (b) Find the potential AV,. (E=8.85×10-12 C²/N×m²) A/2 AV1 K2 A/2 A d K2 KI K1 A/2 2d/3 d/3 K2 A/2 d. AV = 24 V -------
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