Physics M.C.Q (1 Marks)

Current leads by $90^\circ .$

Key concept: Equation for i and $V$: Alternating current or voltage varying as sine function can be written as
$\text{i}=\text{i}_0\ \sin\omega\text{t}=\text{i}_0\ \sin2\pi\text{v t}=\text{i}_0\sin\frac{2\pi}{\text{T}}\text{t}$

and $\text{V}=\text{V}_0\ \sin\omega\text{t}=\text{V}_0\ \sin2\pi\text{v t}=\text{V}_0\sin\frac{2\pi}{\text{T}}\text{t}$
where $i$ and $V$ are instantaneous values of current and voltage,
$i _0$ and $V _0$ are peak values of current and voltage
$\omega =$ Angular frequency in $\frac{\text{red}}{\text{sec}}, v =$ Frequency in $Hz$ and $T =$ time period
Accoeding to the problem, $f = 50Hz, I = 5A$
$\text{t}=\frac{\text{I}}{300}\text{s}$
$\text{I}_0=\text{Peak value}=\sqrt{2}(\text{I}_\text{rms})=5\sqrt{2}$
$=5\sqrt{2}\text{A}$
From, $\text{I}=\text{I}_0\sin\omega\text{t}=5\sqrt{2}\sin2\pi\text{vt}=5\sqrt{2}\sin2\pi\times50\times\frac{1}{300}$
$=5\sqrt{2}\sin\frac{\pi}{3}=5\sqrt{2}\times\frac{\sqrt{3}}{2}=5\sqrt{\frac{3}{2}}\text{A}$

In a series $LCR$ circuit, resonance occurring at $105 \ H$z . At that time, the potential difference across the $100$ resistance is $40 V$ while the potential difference
across the pure inductor is $30 V$ . The inductance $L$ of the inductor is equal to Current( $( i )=\frac{ V _{ r }}{ R }=\frac{100}{40}=2.5 A$
the inductor value is $V _{ L }= i \times( wL )$
$L =\frac{30}{2.5 \times 2 \times \pi 1 \times 105}=1.2 \times 10^{-4} H$

At steady state inductor behave like short circuit.
$\text{i}=\frac{\text{v}}{\text{r}}=\frac{10}{10}=1\text{A}$

Frequency of the source is remain constant $= 50Hz.$

Inductive reactance or capacitive reactance are the impedance of an $AC$ circuit which has the units of ohms.

Since current lead and lag are same So, circuit is in resonance, so, circuit is purely resistive circuit.
So, $\text{i}=\frac{\text{V}}{\text{R}}$
$=\frac{200}{100}$
$=2\text{A}$

Capacitors contain at least two electrical conductors separated by a dielectric/ insulator and is used to store energy electrostatically between the conductors. It acts like an open circuit and hence acts like an infinite resistance for $DC$ currents.

The given $\text{LCR}$ circuit is in resonance. Inductive reactance magnitude $\text{XL}$​ increases as frequency increases while capacitive reactance magnitude $\text{XC}$​ decreases with the increase in frequency.
At one particular frequency, these two reactances are equal in magnitude but opposite in sign; that frequency is called the resonant frequency fO​ for the given circuit. Hence, at resonance:
$\text{XL ​= XC​}$
So the voltage across rthe unknown inductor is $200V$ as same as the voltage across the capacitor.

Resonant frequency, $\text{w}_\text{o}=\frac{1}{\sqrt{\text{LC}}}=\frac{1}{\sqrt{(10\times10^{-3})(10^{-6})}}=\frac{10^4\text{rad}}{\text{s}}$
New frequency $\omega=(0.9)\omega_\text{o}=9\times\frac{10^3\text{rad}}{\text{s}}$
We have new $\text{X}_\text{L}=\omega\text{L}=9\times10^3\times10\times10^{-3}=90\Omega\text{ and}\text{ X}_\text{C}=\frac{1}{\omega\text{C}}=111.11\text{ohm.}$
Thus we calculate new $Z$ as
$\sqrt{3^2+[90-111.11]^2}$
$=21.32\Omega$
Current amplitude $=\frac{\text{V}_\text{o}}{\text{Z}}=\frac{15}{21.32}=0.7\text{A}$

The resonant frequecy of a circuit is given by $\text{f}=\frac{1}{2\pi\sqrt{\text{LC}}}$
Given $\text{L}=0.1\text{H }\text{C}=5\mu\text{F }\text{R}=5\Omega$
Substituting them in formula for $f$ gives
$f = 225.08 Hz.$

$\text{Z}=\frac{1}{\text{WC}}$
$\text{W}\downarrow\text{Z}\uparrow$

Initially there is no $D.C$ current in inductive circuit and maximum $D.C$ current is in capacitive current. Hence, the current is zero in $A _2$ and maximum in $A _1$

In an $a.c.$ circuit containing resistance onIy voltage $\ $ current remain in the same phase. If circuit contains inductance only, voltage remains ahead of current by phase difference of $90^\circ $. If circuit contains capacitance only, current remains ahead of voltage by a phase difference of $90^\circ .$

$\text{f}=\frac{1}{\sqrt{\text{LC}}}$
$\text{or}{\sqrt{\text{LC}}}=\text{T}$

We know that $VR​$ does not depend on driving frequency in a purely resistive circuit. So, If we increase the driving frequency in a circuit with a purely resistive load, then amplitude remain in the same.

Voltage of the source $V = 20$ volts
$\text{Z}=4.33\text{k}\Omega=4.33\times10^3\Omega$
Thus current in the circuit $\text{I}=\frac{\text{V}}{\text{Z}}$
$\therefore\text{I}=\frac{20}{4.33\times10^3}=4.6\times10^{-3}\text{A}$
$\Rightarrow\text{I}=4.6\text{mA}$

We need to find the average potential difference across the mixer. Here by average we mean average over a long period of time. As we know in one complete cycle, average voltage across the mixer is zero. $($In one complete cycle current changes the direction and net voltage across a resistor is zero$).$
So when in one complete cycle voltage drop across the resistor is zero, then the average voltage drop across the resistor $($mixer$)$ is zero.

1. Here, the power factor $\cos\phi\geq0,\text{P}\geq0$.
2. The driving force can give no energy to the oscillator $(P = 0)$ in some cases.
3. The driving force cannot syphon out $(P < 0)$ the energy out of oscillator.

1. It may be defined as cosine of the angle of lag of lead $($i.e., $\cos\phi).$
2. It is also defined as the radio of resistance and impedange $\Big(\text{i.e.,} \frac{\text{R}}{\text{Z}}\Big)$.
3. $\text{The radio}=\frac{\text{True power}}{\text{Apparent power}}=\frac{\text{W}}{\text{VA}}=\frac{\text{kW}}{\text{kVA}}=\cos\phi$

1. As power factor, $\cos\phi=\frac{\text{R}}{\text{Z}}$

2. When $\phi=\frac{\pi}{2} ($in case of $L $ of $C), P = 0.$
3. From $(a)$, it is clear that $P < 0$ is not possible.

Impedance$(Z) =\sqrt{\text{R}^2+\text{X}^2}$
where, $R =$ resistance
$X =$ reactance
Given $\text{R}=4\Omega\text{ and }\text{X}=3\Omega$
Substitute values back in equation
$\text{Z}=\sqrt{4^2+3^2}$
$\text{Z}=5\Omega$

At resonant frequency
$\text{X}_\text{L}=\text{X}_\text{C}=\Big(\omega\text{L}=\frac{1}{\omega\text{C}}\Big)$
At frequencies higher than resonance frequencies
$\text{X}_\text{L}>\text{X}_\text{C}$
i.e., behaviour is inductive.

$V_{Z Y}=V C-V L=0$

$IC$ is $90^{\circ}$ ahead of the applied voltage and $I_L$ lags behind the applied voltage by $90^{\circ}$.
So, there is a phase difference of $180^{\circ}$ between $IL$ and $IC$
$\therefore I=I_C-I_L=0.2 A$

$\text{E}_\text{rms}=\sqrt{\text{V}^2_\text{R}+(\text{V}_\text{L}-\text{V}_\text{c})^2}$
$=\sqrt{40^2+(40-80)^2}$
$=\sqrt{40^2+40^2}$
$=40\sqrt{2}\text{V}$

Resonant frequency $=\frac{1}{2\pi\sqrt{\text{LC}}}$
$=\frac{1}{2\pi\sqrt{10^{-6}}}$
$=\frac{10}{2\pi}\text{Hz}$

The voltages across different elements in $A C$ circuit add vectorially.
The voltages across inductor and capacitor are out of phase and at $90^{\circ}$ to that across resistor.
Hence net voltage across source $=\sqrt{\left( V _1- V _2\right)+ V _3^2}$.

As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence. the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied $AC$ voltage appears across the lnductor, leaving less voltage across the bulb.Therefore' the brightness of the light bulb decreases.

Inductance of the solenoid $=\text{L}=\frac{\mu\text{N}^2\text{A}}{1}=0.493\text{H}$
reactance of this solenoid $=\text{L}\omega=\text{L}2\pi\text{f}=155\Omega$

Since the voltage across the resistive element is same as the voltage applied, the voltage drop across $BC$ is zero. This is possible only when the ohmic value of the element connected across $BC$ is zero. So, $X$ should be combination of inductance and capacitance at resonance.

General sinusoidal voltage variation is given by:
$\text{V}=\text{V}_0\sin(\omega\text{t})$
Here, in this equation $V_0$​ is peak voltage.
So, on comparing both equation we get:
$\text{V}_0=200\text{V},\omega=314$

$\text{I}=\frac{220}{25+\text{j}\Big(0.1\times2\pi\text{f }-\frac{1}{10^{-5}2\pi\text{f}}\Big)}$
$\text{I}=0\text{ at }\text{f}=0$
$\text{I}=0\text{ at }\text{f}=\infty$

Charge on the capacitor at any time $`t'$ is
$\text{q}=\text{CV}(1-\text{e}\frac{-\text{t}}{\tau})$
$\text{at }\text{t}=2\tau$
$\text{q}=\text{CV}(1-\text{e}^{-2})$

Frequency of applied voltage and voltage or current accross any component are same. As we know,
$\text{I}=\frac{\text{V}}{\text{Z}}$ and $Z$ has nothing to do with frequency
$\Rightarrow \text{IandV}$ have same frequency.
And also according to $KVL V = V _1+ V _2+ V _3$
$\Rightarrow $ voltage and current across any component have same frequency i.e $6kHz.$