Embibe Experts Solutions for Chapter: Alternating Current, Exercise 2: Exercise 2

A coil having an inductance of $\frac{1}{\pi } \mathrm{henry}$ is connected in series with a resistance of $300 \mathrm{\Omega }$. If $20 \mathrm{volts}$ from a $200$ cycle source is impressed across the combination, the value of the phase angle between the voltage and the current is,

In series LCR circuit, voltage drop across resistance is $8 \mathrm{volts}$, across inductor is $6 \mathrm{volts}$ and across capacitor is $12 \mathrm{volt}$. Then,

The peak value of the following $\mathrm{A}.\mathrm{C}.$ current, $i=\left[4\sin \left(\omega t\right)+4\sin \left(\omega t+\frac{2\pi }{3}\right)\right] \mathrm{A}$ is,

An alternating voltage $E$(in $\mathrm{volt}$)$=200\sqrt{2}\sin \left(100t\right)$ is connected to a $1 \mathrm{\mu F}$ capacitor through an AC ammeter. The reading of the ammeter shall be,

The equation of alternating current is, $i=50\sqrt{2}\sin \left(400\pi t\right) \mathrm{A}$. Then, the frequency and root-mean-square value of current are, respectively,

$\frac{2.5}{\pi } \mathrm{\mu F}$ capacitor and a $3000 \mathrm{\Omega }$ resistance are joined in series to an $\mathrm{a}.\mathrm{c}.$ source of $200 \mathrm{volt}$, and $50 {\sec }^{-1}$ frequency. The power factor of the circuit and the power dissipated in it will respectively

A coil of resistance $200 \mathrm{ohms}$ and self-inductance $1.0 \mathrm{H}$ has been connected to an A.C. source of frequency $\frac{200}{\pi } \mathrm{Hz}$. The phase difference between voltage and current is 

In an $\mathrm{ac}$ circuit, $V$ and $I$ are given by, $V=100\sin \left(100t\right) \mathrm{volt}$, $I=100\sin \left(100t+\frac{\pi }{3}\right) \mathrm{mA}$. The power dissipated in the circuit is