Phase and Impedance, Reactance, Admittance and Susceptance Questions in English

(A) Reactance is defined as the opposition to the flow of alternating current in an electrical circuit,caused by inductance or capacitance.
Since reactance is a form of resistance to current flow,it is measured in the same unit as resistance.
The $SI$ unit of resistance is the Ohm $(\Omega)$.
Therefore,the unit of reactance is Ohm.

(B) coil possesses both resistance $R$ and inductance $L$.
For $dc$, the frequency $f = 0$, so the inductive reactance $X_L = 2\pi fL = 0$. Thus, the effective resistance is simply $R$.
For $ac$, the frequency $f > 0$, so the inductive reactance $X_L = 2\pi fL$ is non-zero.
The effective resistance (impedance) of the coil in an $ac$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$.
Since $X_L^2 > 0$, it follows that $Z > R$.
Therefore, the effective resistance of the coil increases in an $ac$ circuit.

(C) The impedance $Z$ of an $ac$ series circuit containing a resistance $R$ and a reactance $X$ is given by the formula:
$Z = \sqrt{R^2 + X^2}$
Given:
Resistance $R = 8\,\Omega$
Reactance $X = 6\,\Omega$
Substituting the values into the formula:
$Z = \sqrt{(8)^2 + (6)^2}$
$Z = \sqrt{64 + 36}$
$Z = \sqrt{100}$
$Z = 10\,\Omega$
Therefore,the impedance of the circuit is $10\,\Omega$.

(C) In a series $LCR$ circuit,the power factor is given by $\cos \phi = \frac{R}{Z}$.
Given,$R = 10 \Omega$ and $Z = 20 \Omega$.
Substituting the values,we get $\cos \phi = \frac{10}{20} = \frac{1}{2}$.
Since $\cos \phi = \frac{1}{2}$,the phase difference $\phi = \cos^{-1}(\frac{1}{2}) = 60^o$.
Therefore,the correct option is $C$.

(C) The relationship between phase difference $\phi$ and time difference $\Delta t$ is given by $\Delta t = \frac{\phi}{\omega} = \frac{\phi}{2\pi f}$.
Given: phase difference $\phi = \pi/4$ and frequency $f = 50\, Hz$.
Substituting the values: $\Delta t = \frac{\pi/4}{2\pi \times 50}$.
$\Delta t = \frac{\pi}{4 \times 2\pi \times 50} = \frac{1}{8 \times 50} = \frac{1}{400}\, s$.
$\Delta t = 0.0025\, s = 2.5\, ms$.

(B) Given: $i = 100 \sin(314t) \ A$ and $e = 200 \sin(314t + \pi/3) \ V$.
Comparing with standard equations $i = i_0 \sin(\omega t)$ and $e = e_0 \sin(\omega t + \phi)$,we get peak current $i_0 = 100 \ A$ and peak voltage $e_0 = 200 \ V$.
The impedance $Z$ of the circuit is given by $Z = e_0 / i_0 = 200 / 100 = 2 \ \Omega$.
We know that the impedance is related to resistance $R$ and reactance $X$ by the formula $Z^2 = R^2 + X^2$.
Given $R = 1 \ \Omega$,we have $2^2 = 1^2 + X^2$.
$4 = 1 + X^2 \implies X^2 = 3$.
Therefore,the reactance $X = \sqrt{3} \ \Omega$.

(A) The characteristic impedance $Z$ of a transmission line is given by the formula $Z = \sqrt{\frac{L}{C}}$.
Given:
Inductance $L = 0.40 \mu H = 0.40 \times 10^{-6} H = 4 \times 10^{-7} H$.
Capacitance $C = 1 \times 10^{-11} F$.
Substituting these values into the formula:
$Z = \sqrt{\frac{4 \times 10^{-7}}{1 \times 10^{-11}}}$
$Z = \sqrt{4 \times 10^{4}}$
$Z = 2 \times 10^{2} \Omega = 200 \Omega$.
Therefore,the correct option is $A$.

(D) The frequency of the $AC$ supply is $f = 50\, Hz$.
The time period $T$ of the $AC$ cycle is given by $T = 1/f = 1/50 = 0.02\, s$.
$A$ phase difference of $2\pi$ radians corresponds to one complete time period $T = 0.02\, s$.
Therefore,a phase difference of $\Delta\phi = \pi/4$ radians corresponds to a time difference $\Delta t$ calculated as:
$\Delta t = (\Delta\phi / 2\pi) \times T$
$\Delta t = ((\pi/4) / 2\pi) \times 0.02\, s$
$\Delta t = (1/8) \times 0.02\, s = 0.0025\, s$.
Converting to milliseconds,$\Delta t = 0.0025 \times 1000\, ms = 2.5\, ms$.

(B) real coil possesses both resistance $R$ and self-inductance $L.$
For $DC,$ the inductive reactance $X_L = 2\pi fL$ is zero because the frequency $f = 0.$ Thus,the effective resistance is simply $R = 5\,\Omega.$
For $AC,$ the coil offers an impedance $Z = \sqrt{R^2 + X_L^2},$ where $X_L = \omega L = 2\pi fL > 0.$
Since $Z = \sqrt{R^2 + X_L^2} > R,$ the effective resistance (impedance) of the coil in an $AC$ circuit will be greater than its resistance in a $DC$ circuit.

(A) Impedance $(Z)$ is defined as the effective resistance in an alternating current $(AC)$ circuit.
By Ohm's Law,$Z = \frac{V}{I}$,where $V$ is the potential difference and $I$ is the current.
The dimensional formula for potential difference $(V)$ is $[M L^2 T^{-3} I^{-1}]$.
The dimensional formula for current $(I)$ is $[I]$.
Therefore,the dimension of impedance is:
$[Z] = \frac{[M L^2 T^{-3} I^{-1}]}{[I]} = [M L^2 T^{-3} I^{-2}]$.

(A) phasor is a rotating vector that represents a sinusoidal alternating quantity (like voltage or current) in the complex plane.
It has a length equal to the amplitude of the quantity and rotates at an angular frequency $\omega$ equal to the angular frequency of the alternating quantity.
The projection of the phasor on the vertical axis represents the instantaneous value of the quantity at any time $t$.

(N/A) Capacitive reactance is the opposition offered by a capacitor to the flow of an alternating current $(AC)$.
It is denoted by $X_C$.
The formula for capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$, where $\omega$ is the angular frequency, $f$ is the frequency of the $AC$ source, and $C$ is the capacitance of the capacitor.
The $SI$ unit of capacitive reactance is the ohm $(\Omega)$.

(A) Capacitive reactance $(X_C)$ is defined as the opposition offered by a capacitor to the flow of alternating current.
The formula for capacitive reactance is $X_C = \frac{1}{\omega C}$,where $\omega$ is the angular frequency and $C$ is the capacitance.
Since it represents a form of resistance to current flow,its $SI$ unit is the same as that of resistance,which is the Ohm $(\Omega)$.

(SIEMENS) The term $\omega C$ represents the capacitive susceptance $(B_C)$ in an alternating current circuit.
In an $AC$ circuit,the capacitive reactance is given by $X_C = \frac{1}{\omega C}$.
The unit of capacitive reactance $(X_C)$ is Ohm $(\Omega)$.
Therefore,the unit of $\omega C$ is the reciprocal of Ohm,which is $\Omega^{-1}$ or Siemens $(S)$.

(N/A) $1$. Inductive Reactance $(X_L)$: It is the opposition offered by an inductor to the flow of alternating current. Its formula is $X_L = \omega L = 2\pi f L$,where $\omega$ is the angular frequency,$f$ is the frequency,and $L$ is the inductance. The $SI$ unit is Ohm $(\Omega)$.
$2$. Capacitive Reactance $(X_C)$: It is the opposition offered by a capacitor to the flow of alternating current. Its formula is $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$,where $C$ is the capacitance. The $SI$ unit is Ohm $(\Omega)$.

(B) In an $LCR$ $a.c.$ circuit,the total effective opposition to the flow of current is known as impedance,denoted by $Z$.
By definition,the reciprocal of impedance $(1/Z)$ is known as admittance,denoted by $Y$.
Therefore,$Y = 1/Z$.
Thus,the correct option is $B$.

(B) Inductive reactance $(X_L)$ is defined as the opposition offered by an inductor to the flow of alternating current.
It is given by the formula $X_L = \omega L = 2\pi f L$.
Since it represents the opposition to current flow in an $AC$ circuit, it is analogous to resistance.
Therefore, the $SI$ unit of inductive reactance is the same as that of resistance, which is the ohm $(\Omega)$.

(D) In an $A.C.$ circuit containing a coil (inductor with resistance),the phase difference $\phi$ between the voltage and the current is given by the formula: $\tan \phi = \frac{X_L}{R}$.
Given that the reactance $X_L = \sqrt{3} R$.
Substituting this value into the formula:
$\tan \phi = \frac{\sqrt{3} R}{R} = \sqrt{3}$.
Therefore,the phase difference is $\phi = \tan^{-1}(\sqrt{3})$.

(A) The term $j \omega L$ represents inductive reactance $(X_L)$,which has the same dimensions as resistance $(R)$.
Resistance is defined as $R = \frac{V}{I}$.
Since $V = \frac{W}{Q}$ (where $W$ is work/energy and $Q$ is charge) and $I = \frac{Q}{t}$ (where $t$ is time),
$R = \frac{W/Q}{Q/t} = \frac{W \cdot t}{Q^2}$.
The dimensional formula for work $(W)$ is $[M^1 L^2 T^{-2}]$.
Substituting the dimensions: $R = \frac{[M^1 L^2 T^{-2}] \cdot [T^1]}{[Q^2]} = [M^1 L^2 T^{-1} Q^{-2}]$.
Therefore,the correct option is $A$.

(C) The inductive reactance is given by $X_{L} = \omega L$.
The capacitive reactance is given by $X_{C} = \frac{1}{\omega C}$.
The ratio of inductive reactance to capacitive reactance is:
$\frac{X_{L}}{X_{C}} = \frac{\omega L}{\frac{1}{\omega C}} = \omega L \cdot \omega C = \omega^{2} LC$.

(D) In an $LCR$ series circuit,the impedance $Z$ is given by the formula:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Given values are:
Resistance $R = 4 \ \Omega$
Capacitive reactance $X_C = 6 \ \Omega$
Inductive reactance $X_L = 9 \ \Omega$
Substituting these values into the formula:
$Z = \sqrt{4^2 + (9 - 6)^2}$
$Z = \sqrt{16 + (3)^2}$
$Z = \sqrt{16 + 9}$
$Z = \sqrt{25}$
$Z = 5 \ \Omega$
Therefore,the impedance of the circuit is $5 \ \Omega$.