Physics M.C.Q (1 Marks)

In India, the frequency of $AC$ voltage is $50 Hz.$
It means $50$ waves will be produced in $1 s.$
In one wave, the direction is changed $2$ times.
Thus, in $50$ waves, the direction will be altered $50 \times 2 = 100$ times.
i.e. the direction is changed $100$ times in $1 s.$
Thus the direction is changed in every $\frac{1}{100}\text{second}$

Key Concept: Resonant frequency $($Natural frequency$)$
At resonance $\text{X}_\text{L}=\text{X}_\text{C}\Rightarrow\ \omega_0\text{L}=\frac{1}{\omega_0\text{C}}$
$\Rightarrow\ \omega_0=\frac{1}{\sqrt{\text{LC}}}\frac{\text{red}}{\sec}$
$\Rightarrow\ \text{v}_0=\frac{1}{2\pi\sqrt{\text{LC}}}\text{Hz}$
Resonant frequency in an $L-C-R$ circuit is given by
$\text{v}_0=\frac{1}{2\pi\sqrt{\text{LC}}}$
If $L$ or $C$ increases, the resonant frequency will reduce.
To increase capacitance, we must connect another capacitor parallel to the first.

$\text{V}=\sqrt{\text{V}^2_\text{R}+(\text{V}_\text{L}-\text{V}_\text{C})^2}$
$\text{V}=\sqrt{30^2+(80-40)^2}$
$\text{V}=50\text{ Volt}$

We know at resonance, reactance $($resistance due to inductor and capacitor$)$ be zero $(0).$
At resonance, Impedance $(Z) =$ Resistance $(R)$
Therefore, $Z = 20$ ohm

Supply voltage of an $\text{LCR}$ circuit
$\text{V}=\sqrt{\text{V}^2_\text{R}+(\text{V}_\text{L}-\text{V}_\text{C})^2}$
since inductor and capacitor potentials are out of phase with each other
$=\sqrt{40^2+(60-30)^2\text{V}}$
$=50\text{V}$

$\text{Z}=\sqrt{\text{R}^2+(\text{X}_\text{L}-\text{X}_\text{C})^2}$
$=\sqrt{(10)^2+(8-6)^2}$
$=\sqrt{10^2+2^2}$
$=\sqrt{100+4}$
$=10.2\Omega$

Reactance is the nonresistive component of impedance in an $AC$ circuit, arising from the effect of inductance or capacitance or both and causing the current to be out of phase with the electromotive force causing it.
Therefore, reactance of the $L - C - R$ circuit is $X _{ L }- x _{ C }$

According to the problem, $X _{ L }=1 \Omega, R =2 \Omega$,
$E _{ rms }=6 V, P _{ av }=?$
Average power dissipated in the circuit
$\text{P}_\text{av}=\text{E}_\text{rms}\text{I}_\text{rms}\cos\phi \ .....(\text{i})$
$\text{I}_\text{rms}=\frac{\text{E}_\text{rms}}{\text{Z}}$
$\text{Z}=\sqrt{\text{R}^2+\text{X}^2_\text{L}}$
$=\sqrt{4+1}=\sqrt{5}$
$\text{I}_\text{rms}=\frac{6}{\sqrt{5}}\text{A}$
$\cos\phi=\frac{\text{R}}{\text{Z}}=\frac{2}{\sqrt{5}}$
$\text{P}_\text{av}=6\times\frac{6}{\sqrt{5}}\times\frac{2}{\sqrt{5}}\ \ [\text{from Eq. (i)]}$
$=\frac{72}{\sqrt{5}\sqrt{5}}=\frac{72}{5}=14.4\text{W}$

Current in the circuit is wattless,
Because power $=i^2 R$, if $R=0$, then $P=0$.

$\text{X}_\text{C}=\frac{1}{\omega\text{c}}$
$\therefore\text{X}_\text{C}\propto\frac{1}{\omega}$

$\tan(45)=1\frac{\text{L}\omega}{\text{R}}$
$\text{L}=\frac{\text{R}}{\omega}=\frac{\text{R}}{(2\pi1000)}=.016\text{H}$

$\text{Z}_\text{L}=\text{WL}\ \ \ \text{Z}_\text{C}=\frac{1}{\text{WC}}$
$\text{w}\uparrow,\text{Z}_\text{L}\uparrow\text{Z}_\text{c}\downarrow$

$\text{impedance}×\cos\theta = \text{resistance}$
$\text{impedance} = \frac{\text{resistance}}{\cos\theta}$
$=\frac{\sqrt{300}}{\frac{\cos\pi}{6}}$
$20\Omega$

Resonant frequency in the series $\text{RLC}$ circuit $\text{v}_\text{r}=\frac{1}{2\pi\sqrt{\text{LC}}}$
$\Rightarrow\text{v}_\text{r}\propto\frac{1}{\sqrt{\text{C}}}$
Thus resonant frequency of the circuit increases if the value of $C$ decreases.

In the above image waveform of current and voltage in puerly inductive circuit with time is shown.
It is clear from the image that current lags voltage by $90^\circ .$
Hence phase angle between current and voltage in purely inductive circuit is $\frac{\pi}{2}$

The current in inductor and capacitor is always at an phase difference of $180^\circ $
for $V =\text{V}_\text{o}\sin\omega\text{t}$
Capacitor, $\text{i}=\text{i}_\text{o}\sin(\omega\text{t}-\frac{\pi}{2})$
Capacitor, $\text{i}=\text{i}_\text{o}\sin(\omega\text{t}+\frac{\pi}{2})$
So, the current from both branches will be $0.9 + (-0.4) = 0.5A$

In the problem $\text{XC}=4\Omega$ and $\text{XL}=4\Omega$
So, $V$ across $XC$​ and $XL$​ will be same and in opposite direction, So net voltage will be zero. Since voltmeter is connected parallel to Capacitor and inductor so, it will read $0$ volts.
Current $=\frac{\text{V}}{\text{impedance}}$
$Z=R$ as $X_L=X_C$
Current $=\frac{90}{45}=2\text{A}$

Given equation, $e =80 \sin 100 \pi t. (i)$
Standard equation of instantaneous voltage given by $E = e_m \sin (\omega t) \ldots \ldots (i)$
Compare $(i)$ and $(ii)$, we get $e _{ m }=80 V$ where $e _{ m }$ is the voltage amplitude.
Current amplitude $I _{ m }=\frac{ e _{ m }}{ Z }$ where $Z =$ impedence
$=\frac{80}{20}=4 A$
$I _{ r . m . s }=\frac{4}{\sqrt{2}}=\frac{4 \sqrt{2}}{2}=2 \sqrt{2}=2.828 A$

Here $X_L=X_C ​= R$
$\Rightarrow\frac{1}{2\pi\text{f}}=(\text{R}+2\pi\text{fL})$
$\Rightarrow\text{C}=\frac{1}{2\pi\text{f}(2\pi\text{fL + R})}$

Capacitive reactance $=\frac{1}{\omega\text{c}}$
$=\frac{1}{2\pi\text{fc}}$
$=\frac{1}{2\pi2\times10^3\times50\times10^{-6}}$
$=\frac{5}{\pi}\Omega$

$\text{Z}=\sqrt{\text{R}^2+(\text{X}_\text{L}-\text{X}_\text{C})^2}$

An alternating current $(AC)$ is an electric current whose magnitude and direction vary, unlike direct current, whose direction remains constant.
The usual waveform of an $AC$ power circuit is a sine wave, because this leads to the most efficient transmission of energy. The sine wave oscillates periodically between positive and negative direction.

Effective current is the rms value. Here, $220V$ is the labelled value of $AC$ which is also the rms value. Hence,
$\text{I}_\text{rms}=\frac{\text{E}_\text{rms}}{\text{R}}$
$\text{I}_\text{rms}=\frac{220}{100\times10^3}$
$\text{I}_\text{rms}=2.2{\text{mA}}$

Impedance in branch containing bulb$1$
$\text{Z}_1=200\Omega$
impedance in branch containing bulb$2$
$\text{Z}_2\sqrt{\text{R}^2+\text{X}_\text{L}}^2$
$\text{Z}_2=\sqrt{100^2+100^2}$
$\text{Z}_2=100\sqrt{2}$
Since,
$\text{Z}_1>\text{Z}_2$
$B_2$​ will glow more than $B_1$

This is very fundamental. If we apply separate voltages across resistance and inductor, then in resistance, current and voltage both are in same phase whereas in inductor, current across it lags p.d across it by $\frac{\pi}{2}.$
Now, when we apply voltage across inductor and resistance connencted in series then current through both of them will be same because of $\text{KCL.}$ therefore voltage across resistor will be in same phase with current whereas voltage across inductor will lead the current across it by $\frac{\pi}{2}.$ therefore current and voltage across resistor lags voltage across inductor by $\frac{\pi}{2}$

A constant current exists in a resistor is rms current it is equal to $2.8Amp.$

Watt$-$less power in an $AC$ circuit is basically power supposed to be generated by inductive and capacitive reactance and since they are not resistor they generate any heat, and these power wasted is called watt$-$less power and its phase angle is always $90^{\circ}$ as it has only capacitor and inductor.

In India, the $AC$ mains supply is referred to as single$-$phase alternating current and corresponds to a voltage of $230 V$ at a frequency of $50Hz$, similar to most European countries. Whereas in the $\text{USA, AC}$ mains supply uses $60Hz.$

Capacitor reactance is given by: $\text{X}_\text{c}=\frac{1}{2\pi\text{fC}}$
C is the capacitance.
$X_c\ \&$ f are inversely proportional.

The impedance in $R-L$ circuit is
$\text{Z}=\sqrt{\text{R}^2+{\text{X}^2_\text{L}}}=\sqrt{\text{R}^2+{(\omega\text{L})^2}}$
The current $\text{I}_\text{V}=\frac{\text{E}_\text{V}}{\text{Z}}$
$\frac{\text{E}_\text{v}}{\sqrt{\text{R}^2+\omega^2\text{L}^2}}$

1. $\text{V}=\text{V}_0\sin\omega\text{t}$

Quality factor $(Q)$ of an $L-C-R$ circuit is given by, $\text{Q}=\frac{1}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}$
Tuning of an $L-C-R$ circuit depends on quality factor of the circuit. Tuning will be better when quality factor of the circuit is high.
As, quality factor $(Q)$ of an $L-C-R$ circuit is given by, $\text{Q}=\frac{1}{\text{R}}\sqrt{\frac{\text{L}}{\text{C}}}$
For $Q$ to be high, $R$ should be low, $L$ should be high and $C$ should be low, Therefore option $(c)$ is most apporopriate.

In phasor, $\text{V}_\text{R}=20\angle0;\text{V}_\text{L}=16\angle(-90)$
Total potential difference is $=\text{V}_\text{R}+\text{V}_\text{L}=25.6\angle(-38.66)$
Magnitude of total potential difference $= 25.6V$

We know frequency is given by:
$\text{n}_1=\frac{1}{2\pi\sqrt{\text{LC}_1}}$
And
$\text{n}_2=\frac{1}{2\pi\sqrt{\text{LKC}_1}}$
Thus we get the ratio of frequencies as
$\frac{\text{n}_1}{\text{n}_2}=\frac{150000}{149900}=\sqrt{\text{k}}$
Solving the above equation we get, $\text{K}\approx1.0012$

In an $AC$ circuit, power loss can be minimized by decreasing in resistance and by increasing in inductance.

In a purely capacitive circuit, current leads the voltage by $-\frac{\pi}{2}$

Characteristic impedance of a coaxial cable is $\text{Between }50\Omega \text{ to } 70\Omega$

Circuit is at resonance $(VL = VC)$
$\therefore$ circuit is purely resistance
Resistance is doubled, current in the circuit is half the initial value
$\therefore$ New current $\text{I′}=\frac{\text{I}}{2}$
$\therefore VR = 20V ($equal to applied voltage earlier$)$
$VL = 10V$
$VC = 10V$