(For Nos. 1 and 2): For the circuit shown below, assume all charges and currents are zero at t=0. L= 1H C = 0.25F Switch ON at t = 0 R1 = 4 ohms E = 12V R2 = 6 ohms 1. Solve for the current through R1 in amperes at t=0.5s. Express your answer in four (4) decimal places.

(For Nos. 1 and 2): For the circuit shown below, assume all charges and currents are zero at t=0. L= 1H C = 0.25F Switch ON at t = 0 R1 = 4 ohms E = 12V R2 = 6 ohms 1. Solve for the current through R1 in amperes at t=0.5s. Express your answer in four (4) decimal places.

EBK ELECTRICAL WIRING RESIDENTIAL

19th Edition

ISBN:9781337516549

Author:Simmons

Publisher:Simmons

Chapter27: Service-entrance Equipment

Section: Chapter Questions

Problem 44R

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Concept explainers

Dielectric Breakdown

A dielectric is a material with low electrical conductivity and so is classified as an insulator. Dielectric materials are used to block the flow of electricity through them, allowing them to behave as an insulators. They can also be used to store electric charge or improve capacitance.Dielectrics have no free electrons for conduction because electrons are tightly bound to the nucleus. Because dielectrics contain a small number of electrons for electrical conductivity, they have a dipole. The two primary roles of dielectrics in electrical devices are insulation and electrical energy storage. An electric field polarizes the dielectric, and the polarization of the molecules allows dielectric materials to store charge.

Question

(For Nos. 1 and 2): For the circuit shown below, assume all charges and currents are zero at t=0.
L= 1H
C = 0.25F
Switch ON
at t = 0
R1 = 4 ohms
E = 12V
R2 = 6 ohms
1. Solve for the current through R1 in amperes at t=0.5s. Express your answer in four (4) decimal places.

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