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A 250 Ω resistor is connected in series with a 4.80 µF capacitor and an ac source. The voltage across the capacitor is vC = (7.60 V) sin [(120 rad/s) t]. (a) Determine the capacitive reactance of the capacitor. (b) Derive an expression for the voltage vR across the resistor.

Short Answer

Expert verified

a) The capacitive reactance of the capacitor is 1736 Ω

b) The expression of voltage drop across resistor as a function of time is given byvR=(1.10V)cos[(120rad/s)t]

Step by step solution

01

Concept

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it. When an inductor is attached to an AC supply, the resistance produced by it is called inductive reactance (XL).

Resistance is measure of opposition to the flow of current in a closed electrical circuit. It is measured in Ohm (Ω).

02

Given values

Series Resistance, R = 250 Ω

The capacitance of capacitor, C = 4.80 µF

03

Determination of Capacitive reactance

Voltage across a capacitor is given by

vC=IωCsin(ωt)

Voltage equation given is

vC=(7.6V)sin(120t)

On comparing both equation we get,

localid="1664169758042" IωC=7.6,andω=120rad/s

localid="1664169761543" I=(7.6V)ωC=(7.6V)(120rad/s)(4.80*10-6F)=4.378*10-3A

We know that, localid="1664169766970" V=IXC

So, the capacitive reactance is given by

localid="1664169770785" XC=VI=7.6V4.378*10-3A=1736Ω

Therefore, the capacitive reactance of the capacitor is 1736 Ω

04

Expression of VR as a function of time

Expression of voltage across resistor is given by

vR=VRcoswt

It can be manipulated using VR = IR as

vR=IRcosωtvR=4.378*10-3250Ωcos120tvR=1.10Vcos120rad/st

Therefore, the expression of voltage drop across resistor as a function of time is given byvR=(1.10V)cos[(120rad/s)t]

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