Physics M.C.Q (1 Marks)

$\frac{\text{f}_\text{max}}{\text{f}_\text{max}}=3$
$\therefore\frac{\sqrt{\text{LC}_\text{max}}}{\sqrt{\text{LC}_\text{min}}}=3$
$\frac{\text{C}_\text{max}}{\text{C}_\text{min}}=9$
${\text{C}_\text{min}}=\text{C}$
$\therefore{\text{C}_\text{max}}=\text{9C}$

When the frequency of the source voltage decreases, the impedance of a parallel $RC$ circuit will increase.

Potential difference across the circuit
$\text{V}=\sqrt{\text{V}^2_\text{R}+\text{V}^2_\text{L}}=\sqrt{20^2+16^2}=\sqrt{656}=25.6\text{V}$

$\text{i}=\text{i}_\text{o}\sin(\omega\text{t}+\frac{\pi}{2})$
current leads voltage by $\frac{\pi}{2}$ i.e., current and voltage differ in phase by $90^\circ $

$\text{V}=\sqrt{\text{V}_\text{R}^2+(\text{V}_\text{L}-\text{V}_\text{C})^2}=\sqrt{30^2+(60-20)^2}=50\text{V}$

The power cables have some resistance.
Power lost in the wires can be calculated as $P=I^2 R$ with $R$ as the resistance of the wires and $I$ as the current that passes through them.
Power at the load is $P = VI$.
From this one can see that if voltage is increased by say $n$ times, then only $\frac{1}{n}$ the current is required to deliver the same power. However, if $\frac{1}{ n }$ current is passed
on the same wires, only $\frac{1}{ n ^2}$ of the power will be lost.

At resonance the voltage across $L$ and $C$ are same but opposite.
So, at resonance $\mid\text{V}_\text{L}\mid=\mid\text{V}_\text{C}\mid$
$\therefore\text{V}_\text{R}=\text{V}_\text{app}=120\text{V}$

$\text{X}=0 ($Given$)$
$\text{X}=\text{X}_\text{L}+\text{X}_\text{C}$
$=\omega\text{L}-\frac{1}{\omega\text{C}}=0$
It is possible that the circuit contains an inductor and a capacitor.

$\text{X}_\text{L}=\omega\text{L}$
$\text{X}_\text{C}=\frac{1}{\omega\text{C}}$
If frequency increases that causes '$X_L$' raction of inductor increases and $'X_C'$ reactance of capacitor decreses.

Current leads voltage by $\frac{\pi}{2}$
$\therefore$ phase difference is $=\frac{\pi}{2}$

$\text{I}_\text{p}=14\text{Amp}$
$\text{I}_\text{rms}=\frac{\text{IP}}{\sqrt{2}}=\frac{14}{\sqrt{2}}$
$=9.9=10\text{Amp}.$

Capacitative reactance is an opposition to the change of voltage across an element.
It is denoted by $XC​$ and is inversely proportional to the signal frequency $(f)$ and the capacitance $C$
$\text{X}_\text{c}=\frac{1}{2\pi\text{fc}}$

Impedance first decreases then increases. At resonance frequency $Z$ is minimum.

For maximum power to be delivered from the generator $($or internal reactance $X_g )$ to the load $($of reactance, $X_{\text {}} ),$ $\Rightarrow X_L+X_g=0 ($the total reactance must vanish$)$
$\Rightarrow X_L=-X_g$

Electric charge in alternating current $(AC)$ changes direction periodically. The voltage in $AC$ circuits also periodically reverses because the current changes direction.

$\text{V}=\sqrt{\text{V}_\text{R}^2+(\text{V}_\text{C}-\text{V}_\text{L})^2}=10\text{V}$
$V_C > V_L$​, hence current leads the voltage.
Power factor $=\cos\phi=\frac{8}{10}=0.8$

$\text{Resonant frequency}=\frac{1}{\sqrt{\text{LC}}}$
$\text{LC}\downarrow\text{if }\omega_\text{r}\uparrow$

At resonance $\frac{1}{\omega\text{C}}=\omega\text{L}$
$\therefore$ impedance $= R$

Resonant frequency in series $\text{LCR}$ circuit: $\omega=\sqrt{\frac{1}{\text{LC}}}$
Resonant frequency in series $\text{LCR}$ circuit, $\omega=\sqrt{\frac{1}{\text{LC}}}=\sqrt{\frac{1}{\text{2L}\times\text{2C}}}=\frac{\omega}{2}$

In a purely inductive circuit $($an $AC$ circuit containing inductance only$)$ the current lags behind the voltage by $\frac{\pi}{2}$

Inductors and capacitors bring a phase difference between the voltage and current in the circuit, hence changing the $p.f.$ When only a resistance is present, Poer factor $= 1.$
The power loss in an $AC$ circuit $= E _{ rms } I _{ rms }$ ​Power factor.

For series $C - R$ circuit, the impedance $\text{Z}=\sqrt{\text{R}^2+\text{X}^2_\text{C}}$ where $\text{X}_\text{C}=\frac{\text{i}}{\omega\text{C}}$ and current $\text{I}=\frac{\text{V}}{\text{Z}}$
When the capacitor is filled by mica, the capacitance will be increased. If $C$ increases, $X_C$ decreases, so the current will increase and
hence voltage across resistance increases and voltage across capacitor decreases. thus, $V_a > V_b$

In an $LC$ circuit current oscillates between, maximum and minimum value. So, $LC$ circuit needs oscillations $($electrical$)$. It occurs due to discharging and charging of capacitor and magnetisation and demagnetisation of inductor

$\text{V}^2={\text{V}^2_\text{R}}+{\text{V}^2_\text{c}}$
$20^2=12^2+\text{V}^2_\text{c}$
$\text{V}_\text{c}=\sqrt{20^2-12^2}$
$\text{V}_\text{c}=16\text{ V}$

Resonance frequency f of an $L - C$ circuit can be written as
Resonance frequency $\text{f}=\frac{1}{2\pi\sqrt{\text{LC}}}$ where $L =$ inductance and $C$ is capacitance.

Resonant frequency, $\text{f}_\text{r}=\frac{1}{2\pi\sqrt{\text{LC}}}$
As the frequency is unchanged so $f_r=f_r{ }^{\prime}$
$LC = L ^{\prime} C ^{\prime}= L ^{\prime}(2 C )$
$L =\frac{ L }{2}$

$AC$ voltage or current can be of any form, but the simplest type is a sine wave because any periodic wave can be represented as a combination of sine waves.

$Z-$ Atomic number we study in modern physics
$Z-$ impedance we study in alternating current
$Z-$ zeta potential