Question

A series circuit has a 100-Ω resistor, a 0.100-μF capacitor, and a 2.00-mH inductor connected across a 120-V rms ac source operating at 1000/π Hz. What is the rms voltage across the inductor

Answer 1

Step-by-step explanation:

It is given that,

Resistance of the resistor,
`R=100\ \Omega`

Capacitance of the capacitor,
`C=0.1\ \mu F=0.1* 10^(-6)\ F`

Inductance,
`L=2\ mH=2* 10^(-3)\ H`

rms value of voltage,
`V_(rms)=120\ V`

The Ac source is operating at,
`f=(1000)/(\pi)\ Hz`

The inductive reactance of the inductor is given by :

`X_L=\omega L`

`X_L=2\pi fL`

`X_L=2\pi * (1000)/(\pi)* 2* 10^(-3)`

`X_L=4\ \Omega`

The rms value of current is :

`I_(rms)=(V_(rms))/(R)`

`I_(rms)=(120)/(100)`

`I_(rms)=1.2\ A`

The rms voltage across the inductor is given by :

`V_L=I_(rms)* X_L`

`V_L=1.2\ A* 4\ \Omega`

`V_L=4.8\ V`

So, the voltage across the inductor is 4.8 V. Hence, this is the required solution.