Alternating Current, Voltage (rms and Average) Questions in English

(D) The given current is $i = i_1 + i_2$,where $i_1 = 2\sin(100\pi t)$ and $i_2 = 2\sin(100\pi t + 30^\circ)$.
Using the phasor addition method,the resultant current $i = I_0 \sin(100\pi t + \phi)$.
The amplitude $I_0$ is given by $I_0 = \sqrt{I_{01}^2 + I_{02}^2 + 2I_{01}I_{02}\cos(\theta)}$,where $I_{01} = 2$,$I_{02} = 2$,and $\theta = 30^\circ$.
$I_0 = \sqrt{2^2 + 2^2 + 2(2)(2)\cos(30^\circ)} = \sqrt{4 + 4 + 8(\frac{\sqrt{3}}{2})} = \sqrt{8 + 4\sqrt{3}}$.
$I_0 = \sqrt{4(2 + \sqrt{3})} = 2\sqrt{2 + \sqrt{3}}$.
The $RMS$ value (effective value) $I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2\sqrt{2 + \sqrt{3}}}{\sqrt{2}} = \sqrt{2}\sqrt{2 + \sqrt{3}} = \sqrt{4 + 2\sqrt{3}} = \sqrt{(\sqrt{3}+1)^2} = \sqrt{3} + 1 \, A$.
Since this value is not among the options,the correct choice is $D$.

(B) The $r.m.s.$ value of a current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_0^T i^2 dt}$.
For case $I$ and $II$,the current is a rectified sine wave: $i = |I_0 \sin(\omega t)|$. The $r.m.s.$ value is $I_1 = I_2 = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0$.
For case $III$,the current is a square wave: $i = I_0$ for half cycle and $-I_0$ for the other half. The $r.m.s.$ value is $I_3 = \sqrt{\frac{1}{T} (I_0^2 \cdot \frac{T}{2} + (-I_0)^2 \cdot \frac{T}{2})} = I_0$.
For case $IV$,the current is a triangular wave: $i = \frac{2I_0}{T} t$ for $0 < t < T/2$. The $r.m.s.$ value is $I_4 = \sqrt{\frac{1}{T} \int_0^T i^2 dt} = \frac{I_0}{\sqrt{3}} \approx 0.577 I_0$.
Comparing the values: $I_3 = I_0$,$I_1 = I_2 = 0.707 I_0$,and $I_4 = 0.577 I_0$.
Thus,the correct relation is $I_3 > I_1 = I_2 > I_4$.

(D) Given voltage $e(t) = 5[\cos \omega t + \sqrt{3} \sin \omega t]$.
Multiplying and dividing by $2$,we get $e(t) = 10[\frac{1}{2} \cos \omega t + \frac{\sqrt{3}}{2} \sin \omega t]$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can write $e(t) = 10 \sin(\omega t + \frac{\pi}{3})$.
The current is given as $i(t) = 5 \sin(\omega t + \frac{\pi}{4})$.
The phase of voltage is $\phi_v = \frac{\pi}{3}$ and the phase of current is $\phi_i = \frac{\pi}{4}$.
The phase difference is $\Delta \phi = \phi_v - \phi_i = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$.
Since $\phi_v > \phi_i$,the voltage leads the current by $\frac{\pi}{12}$. However,checking the options,the current leads the voltage by $-\frac{\pi}{12}$ or voltage leads current by $\frac{\pi}{12}$. Given the standard interpretation of such problems,if the current leads by $\frac{\pi}{12}$ is not an option,we re-evaluate: $\phi_i - \phi_v = \frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$. Thus,voltage leads current by $\frac{\pi}{12}$. Since this is not listed,let's re-check the calculation: $\frac{\pi}{3} \approx 60^\circ$,$\frac{\pi}{4} = 45^\circ$. $60^\circ - 45^\circ = 15^\circ = \frac{\pi}{12}$. The correct physical result is voltage leads current by $\frac{\pi}{12}$. Given the provided options,there might be a typo in the question's options,but based on the math,the phase difference is $\frac{\pi}{12}$.

(A) For the given square wave,the voltage $V(t)$ is $V_0$ for $0 < t < T/2$ and $-V_0$ for $T/2 < t < T$.
$1$. Mean value for a half cycle ($0$ to $T/2$):
$V_{mean} = \frac{1}{T/2} \int_0^{T/2} V_0 dt = \frac{2}{T} [V_0 t]_0^{T/2} = \frac{2}{T} \cdot V_0 \cdot \frac{T}{2} = V_0$.
$2$. $rms$ value for a full cycle (or half cycle):
$V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2 dt} = \sqrt{\frac{1}{T} [\int_0^{T/2} V_0^2 dt + \int_{T/2}^T (-V_0)^2 dt]} = \sqrt{\frac{1}{T} [V_0^2 \cdot \frac{T}{2} + V_0^2 \cdot \frac{T}{2}]} = \sqrt{\frac{1}{T} \cdot V_0^2 \cdot T} = V_0$.
Thus,the mean and $rms$ values are $V_0$ and $V_0$ respectively.

(B) The instantaneous voltage in an $ac$ circuit is given by $V = V_0 \sin(\omega t),$ where $\omega = 2\pi n$ and $n$ is the frequency of the source.
The instantaneous power $P$ in the circuit is given by $P = V \cdot I = V_0 \sin(\omega t) \cdot I_0 \sin(\omega t + \phi) = V_0 I_0 \sin(\omega t) \sin(\omega t + \phi).$
Using the trigonometric identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)],$ we get:
$P = \frac{V_0 I_0}{2} [\cos(\phi) - \cos(2\omega t + \phi)].$
Since the power expression contains a term with $2\omega,$ the frequency of the power variation is $2n.$

(D) The voltage waveform is periodic with a period of $T$. Let us consider one cycle from $t = 0$ to $t = T$.
In the interval $t = 0$ to $t = T$,the voltage $V(t)$ varies linearly from $-V_0$ to $+V_0$.
The equation of the line passing through $(0, -V_0)$ and $(T, V_0)$ is given by:
$V(t) = \frac{V_0 - (-V_0)}{T - 0} t - V_0 = \frac{2V_0}{T} t - V_0$
The $RMS$ value $V_{rms}$ is defined as:
$V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2 dt}$
Substituting $V(t)$:
$V_{rms}^2 = \frac{1}{T} \int_0^T \left( \frac{2V_0}{T} t - V_0 \right)^2 dt$
Let $u = \frac{2V_0}{T} t - V_0$,then $du = \frac{2V_0}{T} dt$,so $dt = \frac{T}{2V_0} du$.
When $t=0, u=-V_0$; when $t=T, u=V_0$.
$V_{rms}^2 = \frac{1}{T} \int_{-V_0}^{V_0} u^2 \left( \frac{T}{2V_0} \right) du = \frac{1}{2V_0} \left[ \frac{u^3}{3} \right]_{-V_0}^{V_0} = \frac{1}{2V_0} \left( \frac{V_0^3}{3} - \frac{(-V_0)^3}{3} \right) = \frac{1}{2V_0} \left( \frac{2V_0^3}{3} \right) = \frac{V_0^2}{3}$
Therefore,$V_{rms} = \frac{V_0}{\sqrt{3}}$.

(B) The $r.m.s.$ value of a periodic function is given by $V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2 dt}$.
From the figure,the voltage $V$ is $V_0$ for $0 \le t < \frac{T}{4}$ and $0$ for $\frac{T}{4} \le t < T$.
Substituting these values into the formula:
$V_{rms} = \sqrt{\frac{1}{T} \left( \int_0^{T/4} V_0^2 dt + \int_{T/4}^T 0^2 dt \right)}$
$V_{rms} = \sqrt{\frac{1}{T} \left( V_0^2 [t]_0^{T/4} + 0 \right)}$
$V_{rms} = \sqrt{\frac{1}{T} \cdot V_0^2 \cdot \frac{T}{4}}$
$V_{rms} = \sqrt{\frac{V_0^2}{4}}$
$V_{rms} = \frac{V_0}{2}$

(A) The given waveform is a square wave with amplitude $V_0 = 10 \ V$ and $-V_0 = -10 \ V$.
For a square wave,the instantaneous voltage $V(t)$ is $10 \ V$ for half the period and $-10 \ V$ for the other half.
The $r.m.s.$ voltage is defined as $V_{r.m.s.} = \sqrt{\frac{1}{T} \int_{0}^{T} V^2(t) \ dt}$.
Since $V^2(t) = (10)^2 = 100$ throughout the entire period $T$,we have:
$V_{r.m.s.} = \sqrt{\frac{1}{T} \int_{0}^{T} 100 \ dt} = \sqrt{\frac{1}{T} \cdot 100 \cdot T} = \sqrt{100} = 10 \ V$.
Thus,the correct option is $A$.

(B) The $r.m.s.$ value of a current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2 dt}$.
Given $I = I_0 + I_1 \sin \omega t$,we have $I^2 = I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t$.
Integrating over one complete time period $T = \frac{2\pi}{\omega}$:
$I_{rms}^2 = \frac{1}{T} \int_{0}^{T} (I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t) dt$.
Since the average value of $\sin \omega t$ over a full cycle is $0$,the term $\int_{0}^{T} 2 I_0 I_1 \sin \omega t dt = 0$.
The average value of $\sin^2 \omega t$ over a full cycle is $\frac{1}{2}$.
Thus,$I_{rms}^2 = I_0^2 + I_1^2 \left( \frac{1}{2} \right) + 0 = I_0^2 + \frac{I_1^2}{2}$.
Therefore,$I_{rms} = \sqrt{I_0^2 + \frac{I_1^2}{2}}$.

(C) Given the equation: $V = 100 \sin(100 \pi t) \cos(100 \pi t)$.
Using the trigonometric identity $\sin(2\theta) = 2 \sin \theta \cos \theta$,we can rewrite the equation as:
$V = 50 \times (2 \sin(100 \pi t) \cos(100 \pi t))$
$V = 50 \sin(200 \pi t)$.
Comparing this with the standard form $V = V_0 \sin(\omega t)$,where $V_0$ is the peak voltage and $\omega = 2 \pi f$:
Peak voltage $V_0 = 50 \text{ V}$.
Angular frequency $\omega = 200 \pi \text{ rad/s}$.
Since $\omega = 2 \pi f$,we have $200 \pi = 2 \pi f$,which gives $f = 100 \text{ Hz}$.
Thus,the peak voltage is $50 \text{ V}$ and the frequency is $100 \text{ Hz}$. Therefore,option $C$ is correct.

(D) The given voltage is $e = e_1 \sin \omega t + e_2 \cos \omega t$.
We can rewrite this as $e = E_0 \sin(\omega t + \phi)$,where $E_0 = \sqrt{e_1^2 + e_2^2}$ is the peak voltage.
The root mean square $(RMS)$ value of an alternating voltage $e = E_0 \sin(\omega t + \phi)$ is given by $V_{rms} = \frac{E_0}{\sqrt{2}}$.
Substituting the value of $E_0$,we get $V_{rms} = \frac{\sqrt{e_1^2 + e_2^2}}{\sqrt{2}} = \sqrt{\frac{e_1^2 + e_2^2}{2}}$.

(B) The root mean square $(I_{\text{rms}})$ value of an alternating current is defined as the square root of the mean of the squares of the instantaneous current over one complete cycle.
For a sinusoidal alternating current given by $I = I_{0} \sin(\omega t)$,the $I_{\text{rms}}$ is calculated as:
$I_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} I^{2} dt}$
Substituting $I = I_{0} \sin(\omega t)$:
$I_{\text{rms}} = \sqrt{\frac{1}{T} \int_{0}^{T} I_{0}^{2} \sin^{2}(\omega t) dt}$
$I_{\text{rms}} = I_{0} \sqrt{\frac{1}{T} \int_{0}^{T} \frac{1 - \cos(2\omega t)}{2} dt}$
$I_{\text{rms}} = I_{0} \sqrt{\frac{1}{2T} [t - \frac{\sin(2\omega t)}{2\omega}]_{0}^{T}}$
Since $\sin(2\omega T) = \sin(4\pi) = 0$,we get:
$I_{\text{rms}} = I_{0} \sqrt{\frac{T}{2T}} = \frac{I_{0}}{\sqrt{2}}$
Therefore,the correct relation is $I_{\text{rms}} = \frac{1}{\sqrt{2}} I_{0}$.

(B) The root mean square $(RMS)$ current is defined as $I_{rms} = \sqrt{\frac{1}{T_2 - T_1} \int_{T_1}^{T_2} i^2 dt}$.
Given $i = 2\sqrt{t}$,so $i^2 = 4t$.
The time interval is from $T_1 = 2\,s$ to $T_2 = 4\,s$,so $T_2 - T_1 = 4 - 2 = 2\,s$.
Now,calculate the integral: $\int_{2}^{4} 4t \,dt = [2t^2]_{2}^{4} = 2(4^2 - 2^2) = 2(16 - 4) = 2(12) = 24$.
Substitute these values into the $RMS$ formula:
$I_{rms} = \sqrt{\frac{24}{2}} = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\,A$.

(C) The given equation is $i = a \sin(\omega t) + b \cos(\omega t)$.
We can rewrite this as $i = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin(\omega t) + \frac{b}{\sqrt{a^2 + b^2}} \cos(\omega t) \right)$.
Let $\frac{a}{\sqrt{a^2 + b^2}} = \cos \phi$ and $\frac{b}{\sqrt{a^2 + b^2}} = \sin \phi$.
Then $i = \sqrt{a^2 + b^2} \sin(\omega t + \phi)$.
The peak value (amplitude) of the current is $i_0 = \sqrt{a^2 + b^2}$.
The $rms$ value of an alternating current is given by $i_{rms} = \frac{i_0}{\sqrt{2}}$.
Substituting the value of $i_0$,we get $i_{rms} = \frac{\sqrt{a^2 + b^2}}{\sqrt{2}} = \sqrt{\frac{a^2 + b^2}{2}}$.

(D) The given equation for current is $I = I_0 \sin(\omega t)$,where $I_0 = 100 \, A$ and $\omega = 200 \pi \, rad/s$.
For the current to reach its peak value $(I = I_0)$,the argument of the sine function must be $\frac{\pi}{2}$.
Thus,$\omega t = \frac{\pi}{2}$.
Substituting the value of $\omega$: $200 \pi t = \frac{\pi}{2}$.
Solving for $t$: $t = \frac{\pi}{2 \times 200 \pi} = \frac{1}{400} \, s$.

(C) Given the current equation: $i = 2\sin(100\pi t) + 2\cos(100\pi t + 30^{\circ})$.
Using the trigonometric identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we expand the second term:
$i = 2\sin(100\pi t) + 2[\cos(100\pi t)\cos(30^{\circ}) - \sin(100\pi t)\sin(30^{\circ})]$.
Substitute $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$:
$i = 2\sin(100\pi t) + 2[\cos(100\pi t) \cdot \frac{\sqrt{3}}{2} - \sin(100\pi t) \cdot \frac{1}{2}]$.
$i = 2\sin(100\pi t) + \sqrt{3}\cos(100\pi t) - \sin(100\pi t)$.
$i = \sin(100\pi t) + \sqrt{3}\cos(100\pi t)$.
This is in the form $i = I_m \sin(100\pi t + \phi)$,where the amplitude $I_m = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2\,A$.
The effective value ($RMS$ value) is given by $I_{\text{rms}} = \frac{I_m}{\sqrt{2}}$.
$I_{\text{rms}} = \frac{2}{\sqrt{2}} = \sqrt{2}\,A$.

(B) The given equation for the alternating current is $I = I_0 \cos(\omega t)$.
Comparing this with the standard form $I = I_{peak} \cos(\omega t)$,the peak value of the current is $I_{peak} = I_0$.
The root mean square $(rms)$ value of an alternating current is defined as $I_{rms} = \frac{I_{peak}}{\sqrt{2}}$.
Substituting the value of $I_{peak}$,we get $I_{rms} = \frac{I_0}{\sqrt{2}}$.
Therefore,the $rms$ value is $\frac{I_0}{\sqrt{2}}$ and the peak value is $I_0$.

(B) The $rms$ value of current is defined as $I_{rms} = \sqrt{\frac{1}{T_2 - T_1} \int_{T_1}^{T_2} I^2 dt}$.
Given $I = 2\sqrt{t}$,so $I^2 = 4t$.
The interval is from $T_1 = 2 \, s$ to $T_2 = 4 \, s$,so the duration is $T_2 - T_1 = 4 - 2 = 2 \, s$.
Calculating the integral: $\int_{2}^{4} 4t \, dt = [2t^2]_{2}^{4} = 2(4^2 - 2^2) = 2(16 - 4) = 2(12) = 24$.
Now,the mean square value is $\langle I^2 \rangle = \frac{1}{2} \times 24 = 12 \, A^2$.
Therefore,$I_{rms} = \sqrt{12} = 2\sqrt{3} \, A$.

(B) standard $DC$ voltmeter cannot measure $AC$ voltage because the average value of an alternating voltage over a full cycle is zero, causing the instrument to show a zero reading.
To measure $AC$ voltage, we use an instrument that operates on the heating effect of current, which is independent of the direction of the current.
A hot wire voltmeter works on the principle of the heating effect of current $(H = I^2Rt)$, where the deflection is proportional to the square of the current (or voltage). Therefore, it can measure the $RMS$ value of $AC$ voltage.

(C) The given current is $i = I_0 + I_1 \sin \omega t$,where $I_0 = 3$ and $I_1 = 4$.
The effective $(RMS)$ value of current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_0^T i^2 dt}$.
Substituting $i = I_0 + I_1 \sin \omega t$:
$I_{rms} = \sqrt{\frac{1}{T} \int_0^T (I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t) dt}$.
Using the properties of integration over a full cycle:
$\frac{1}{T} \int_0^T dt = 1$,$\frac{1}{T} \int_0^T \sin^2 \omega t dt = \frac{1}{2}$,and $\frac{1}{T} \int_0^T \sin \omega t dt = 0$.
Thus,$I_{rms} = \sqrt{I_0^2 + \frac{I_1^2}{2}}$.
Substituting the values $I_0 = 3$ and $I_1 = 4$:
$I_{rms} = \sqrt{3^2 + \frac{4^2}{2}} = \sqrt{9 + \frac{16}{2}} = \sqrt{9 + 8} = \sqrt{17}$.

(B) From the graph,the time period $T$ (time taken to complete one full cycle) for wave $M$ is $0.4 \, s$.
Therefore,the frequency $f = \frac{1}{T} = \frac{1}{0.4} = 2.5 \, Hz$.
Wave $M$ starts at $t=0$ with a value of $0$,while wave $N$ reaches the same point at $t=0.1 \, s$.
The time shift $\Delta t$ between the two waves is $0.1 \, s$.
The phase difference $\Delta \phi$ is given by $\Delta \phi = \frac{2\pi}{T} \times \Delta t = \frac{2\pi}{0.4} \times 0.1 = \frac{2\pi}{4} = \frac{\pi}{2}$.
Since wave $N$ lags behind wave $M$,the phase lead of $N$ over $M$ is $-\frac{\pi}{2}$ radians.

(B) The resulting current is $I = a + b \sin \omega t$.
The effective $(RMS)$ value of the current is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2 dt}$.
Squaring both sides,we get $I_{rms}^2 = \frac{1}{T} \int_{0}^{T} (a + b \sin \omega t)^2 dt$.
Expanding the integrand: $I_{rms}^2 = \frac{1}{T} \int_{0}^{T} (a^2 + b^2 \sin^2 \omega t + 2ab \sin \omega t) dt$.
Integrating over one full time period $T$:
$\int_{0}^{T} a^2 dt = a^2 T$
$\int_{0}^{T} b^2 \sin^2 \omega t dt = b^2 \int_{0}^{T} \frac{1 - \cos 2\omega t}{2} dt = \frac{b^2 T}{2}$
$\int_{0}^{T} 2ab \sin \omega t dt = 0$ (since the average of a sine wave over a full cycle is zero).
Substituting these values: $I_{rms}^2 = \frac{1}{T} (a^2 T + \frac{b^2 T}{2} + 0) = a^2 + \frac{b^2}{2}$.
Therefore,$I_{rms} = \sqrt{a^2 + \frac{b^2}{2}}$.

(C) The given current is $i = i_1 \cos \omega t + i_2 \sin \omega t$.
The $r.m.s.$ value of current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_0^T i^2 dt}$.
First,calculate $i^2 = (i_1 \cos \omega t + i_2 \sin \omega t)^2 = i_1^2 \cos^2 \omega t + i_2^2 \sin^2 \omega t + 2 i_1 i_2 \sin \omega t \cos \omega t$.
Integrating over a full time period $T = \frac{2\pi}{\omega}$:
$\int_0^T \cos^2 \omega t dt = \frac{T}{2}$,$\int_0^T \sin^2 \omega t dt = \frac{T}{2}$,and $\int_0^T \sin \omega t \cos \omega t dt = 0$.
Thus,the mean square value is $i_{rms}^2 = \frac{1}{T} [i_1^2 (\frac{T}{2}) + i_2^2 (\frac{T}{2}) + 0] = \frac{i_1^2 + i_2^2}{2}$.
Therefore,$i_{rms} = \sqrt{\frac{i_1^2 + i_2^2}{2}} = \frac{1}{\sqrt{2}} (i_1^2 + i_2^2)^{1/2}$.

(A) The average value of current $I$ over a time interval $[t_1, t_2]$ is given by $I_{av} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} I \, dt$.
Here,$t_1 = 0$ and $t_2 = \frac{\pi}{\omega}$.
$I_{av} = \frac{1}{\frac{\pi}{\omega} - 0} \int_0^{\frac{\pi}{\omega}} I_0 \sin \omega t \, dt$
$I_{av} = \frac{\omega}{\pi} I_0 \left[ \frac{-\cos \omega t}{\omega} \right]_0^{\frac{\pi}{\omega}}$
$I_{av} = \frac{I_0}{\pi} [-\cos(\omega \cdot \frac{\pi}{\omega}) - (-\cos(0))]$
$I_{av} = \frac{I_0}{\pi} [-\cos(\pi) + \cos(0)]$
$I_{av} = \frac{I_0}{\pi} [-(-1) + 1] = \frac{I_0}{\pi} [1 + 1] = \frac{2I_0}{\pi}$.

(D) The instantaneous current is given by $I = I_0 \sin(\omega t)$.
To change from zero to maximum value,the current must go from $I = 0$ to $I = I_0$.
Thus,$\sin(\omega t) = 1$.
This implies $\omega t = \frac{\pi}{2}$.
Substituting $\omega = 2\pi f$,we get $2\pi f t = \frac{\pi}{2}$.
Given frequency $f = 50\,Hz$,the equation becomes $2\pi(50)t = \frac{\pi}{2}$.
$100\pi t = \frac{\pi}{2}$.
$t = \frac{1}{200}\,s = 0.005\,s = 5\,ms$.

(C) The average value of a periodic function is given by the area under the curve over one complete time period divided by the time period.
From the figure,the potential difference $V$ is non-zero only from $t = 0$ to $t = T/2$ within one cycle of duration $T$.
The area under the $V-t$ graph for one cycle is the area of the triangle with base $T/2$ and height $V_0$:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{T}{2} \times V_0 = \frac{V_0 T}{4}$.
The average value $\langle V \rangle$ is:
$\langle V \rangle = \frac{\text{Area}}{T} = \frac{\frac{V_0 T}{4}}{T} = \frac{V_0}{4}$.

(B) $(1)$ For $x = x_0 \sin \omega t$,the $r.m.s.$ value is $I_{rms} = \frac{x_0}{\sqrt{2}}$. Thus,$(A \to ii)$.
$(2)$ For $x = x_0 \sin \omega t \cos \omega t = \frac{x_0}{2} \sin(2\omega t)$,the peak value is $\frac{x_0}{2}$. The $r.m.s.$ value is $\frac{x_0/2}{\sqrt{2}} = \frac{x_0}{2\sqrt{2}}$. Thus,$(B \to iii)$.
$(3)$ For $x = x_0 \sin \omega t + x_0 \cos \omega t = \sqrt{2} x_0 \sin(\omega t + \frac{\pi}{4})$,the peak value is $\sqrt{2} x_0$. The $r.m.s.$ value is $\frac{\sqrt{2} x_0}{\sqrt{2}} = x_0$. Thus,$(C \to i)$.

(B) The given current is $i = i_1 \sin \omega t + i_2 \cos \omega t$.
We can rewrite this as $i = I_0 \sin(\omega t + \phi)$,where $I_0$ is the peak current.
The square of the peak current is $I_0^2 = i_1^2 + i_2^2$.
Thus,$I_0 = \sqrt{i_1^2 + i_2^2}$.
The $r.m.s.$ value of a sinusoidal current is given by $I_{rms} = \frac{I_0}{\sqrt{2}}$.
Substituting the value of $I_0$,we get $I_{rms} = \frac{\sqrt{i_1^2 + i_2^2}}{\sqrt{2}} = \sqrt{\frac{i_1^2 + i_2^2}{2}}$.

(C) The given waveform represents a half-wave rectified $A.C.$ signal.
For one complete cycle of period $T$,the voltage $V(t)$ is defined as:
$V(t) = V_0 \sin(\omega t)$ for $0 \le t \le T/2$
$V(t) = 0$ for $T/2 < t \le T$
where $\omega = \frac{2\pi}{T}$.
The $r.m.s.$ value is defined as $V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} V^2(t) dt}$.
$V_{rms}^2 = \frac{1}{T} \left[ \int_{0}^{T/2} (V_0 \sin(\omega t))^2 dt + \int_{T/2}^{T} 0^2 dt \right]$
$V_{rms}^2 = \frac{V_0^2}{T} \int_{0}^{T/2} \sin^2(\frac{2\pi t}{T}) dt$
Using the identity $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$:
$V_{rms}^2 = \frac{V_0^2}{T} \int_{0}^{T/2} \frac{1 - \cos(\frac{4\pi t}{T})}{2} dt = \frac{V_0^2}{2T} \left[ t - \frac{T}{4\pi} \sin(\frac{4\pi t}{T}) \right]_{0}^{T/2}$
$V_{rms}^2 = \frac{V_0^2}{2T} \left[ (\frac{T}{2} - 0) - (0 - 0) \right] = \frac{V_0^2}{2T} \cdot \frac{T}{2} = \frac{V_0^2}{4}$
Therefore,$V_{rms} = \sqrt{\frac{V_0^2}{4}} = \frac{V_0}{2}$.

(C) The given current is $i = 3 + 4 \sin \omega t$.
To find the $RMS$ value,we first calculate the mean square value $i^2_{rms} = \overline{i^2}$.
$i^2 = (3 + 4 \sin \omega t)^2 = 9 + 16 \sin^2 \omega t + 24 \sin \omega t$.
Taking the average over a full cycle:
$\overline{i^2} = \overline{9} + 16 \overline{\sin^2 \omega t} + 24 \overline{\sin \omega t}$.
We know that $\overline{\sin^2 \omega t} = 1/2$ and $\overline{\sin \omega t} = 0$ over a full cycle.
Therefore,$i^2_{rms} = 9 + 16(1/2) + 0 = 9 + 8 = 17$.
The $RMS$ value is $i_{rms} = \sqrt{\overline{i^2}} = \sqrt{17}$.

(B) The given current wave is $i = 5 + 5 \sin(100 \omega t)$.
To find the average value over one time period $T$,we use the formula $\langle i \rangle = \frac{1}{T} \int_{0}^{T} i \, dt$.
Substituting the expression for $i$:
$\langle i \rangle = \frac{1}{T} \int_{0}^{T} (5 + 5 \sin(100 \omega t)) \, dt$.
This can be split into two parts:
$\langle i \rangle = \frac{1}{T} \int_{0}^{T} 5 \, dt + \frac{1}{T} \int_{0}^{T} 5 \sin(100 \omega t) \, dt$.
The average value of a sine function over one complete cycle is $0$,i.e.,$\langle \sin(100 \omega t) \rangle = 0$.
Therefore,$\langle i \rangle = 5 + 0 = 5 \text{ A}$.

(A) The given $AC$ voltage is the $rms$ value,so $E_{rms} = 220\,V$.
The peak voltage $E_{0}$ is related to $E_{rms}$ by the formula $E_{rms} = \frac{E_{0}}{\sqrt{2}}$,which gives $E_{0} = E_{rms} \times \sqrt{2} = 220\sqrt{2}\,V$.
The average $emf$ over a positive half cycle is given by the formula $E_{avg} = \frac{2}{\pi} E_{0}$.
Substituting the value of $E_{0}$,we get $E_{avg} = \frac{2}{\pi} \times 220\sqrt{2}$.
Using $\pi \approx 3.14$ and $\sqrt{2} \approx 1.414$,we get $E_{avg} = \frac{2}{3.14} \times 220 \times 1.414 \approx 0.637 \times 311.13 \approx 198.18\,V$.
Rounding to the nearest integer,the average $emf$ is $198\,V$.

(B) An alternating current $(AC)$ is defined as a current that periodically reverses its direction and changes its magnitude with time.
In the given graphs,the $EMF$ (and consequently the current) in all four cases $(a), (b), (c),$ and $(d)$ crosses the time axis,meaning the $EMF$ changes its polarity (sign) periodically.
Since the polarity of the $EMF$ changes in all four graphs,all of them represent alternating $EMF$ or $AC$ waveforms.
Therefore,the correct option is $(b)$.

(N/A) Given power $P = 100\;W$ and rms voltage $V = 220\;V$.
The resistance of the bulb is calculated as:
$R = \frac{V^2}{P} = \frac{(220\;V)^2}{100\;W} = 484\;\Omega$
$(b)$ The peak voltage $(V_m)$ of the source is given by:
$V_m = \sqrt{2} \times V = 1.414 \times 220\;V \approx 311\;V$
$(c)$ The rms current $(I)$ through the bulb is calculated using $P = I \times V$:
$I = \frac{P}{V} = \frac{100\;W}{220\;V} \approx 0.454\;A$

(N/A) Given,peak voltage $V_{0} = 300 \; V$.
The $rms$ voltage is given by the relation $V_{rms} = \frac{V_{0}}{\sqrt{2}}$.
Substituting the values: $V_{rms} = \frac{300}{1.414} \approx 212.1 \; V$.
$(b)$ Given,$rms$ current $I_{rms} = 10 \; A$.
The peak current $I_{0}$ is given by the relation $I_{0} = \sqrt{2} \times I_{rms}$.
Substituting the values: $I_{0} = 1.414 \times 10 = 14.14 \; A$.

(N/A) $DC$ signals (current or voltage) do not change their direction with time. They are unidirectional signals.
If the voltage obtained from a source varies like a sine function with time,then such a voltage is called an alternating voltage ($AC$ voltage).
The current driven in a circuit by an $AC$ voltage is called an alternating current ($AC$ current).
$AC$ is preferred over $DC$ for the following reasons:
$1$. $AC$ voltages can be easily and efficiently converted from one voltage level to another using transformers,which is not possible with $DC$.
$2$. Electrical energy can be transmitted and distributed over long distances much more economically using $AC$ compared to $DC$.

(N/A) An $A.C.$ (Alternating Current) signal is an electric current or voltage that periodically reverses its direction and changes its magnitude continuously with time.
Unlike $D.C.$ (Direct Current),which flows in a single direction,$A.C.$ follows a sinusoidal waveform in most power applications.
The mathematical expression for an $A.C.$ voltage is typically given by $V(t) = V_m \sin(\omega t + \phi)$,where $V_m$ is the peak voltage,$\omega$ is the angular frequency,and $\phi$ is the phase constant.
$A.C.$ is widely used for power distribution because its voltage can be easily stepped up or down using transformers.

(N/A) The root mean square (rms) value of an alternating current is defined as the equivalent steady direct current $(DC)$ which,when flowing through a given resistor,produces the same amount of heat in a given time as the alternating current does over a complete cycle.
The rms current is denoted by $I$ or $I_{rms}$.
The relationship between the rms current $I$ and the peak current $I_{m}$ is given by:
$I = \frac{I_{m}}{\sqrt{2}} \approx 0.707 I_{m}$
Mathematically,it is defined as:
$I_{rms} = \sqrt{\overline{I}^{2}} = \sqrt{\frac{1}{T} \int_{0}^{T} I_{m}^{2} \sin^{2}(\omega t) dt} = \sqrt{\frac{1}{2} I_{m}^{2}} = \frac{I_{m}}{\sqrt{2}}$
Similarly,for voltage:
$V = \frac{V_{m}}{\sqrt{2}} \approx 0.707 V_{m}$
These relations show that $V = IR$ holds for $AC$ circuits using rms values,similar to $DC$ circuits.
The graph of current versus $\omega t$ shows the sinusoidal variation of current with peak value $I_{m}$ and the constant rms value $I$ represented as a horizontal line.

(N/A) $AC$ voltage is a voltage that periodically reverses its direction and changes its magnitude continuously with time.
The equation for $AC$ voltage is given by:
$v(t) = V_m \sin(\omega t + \phi)$
Where:
$v(t)$ is the instantaneous voltage at time $t$.
$V_m$ is the peak or maximum voltage (amplitude).
$\omega$ is the angular frequency,where $\omega = 2\pi f$ ($f$ is the frequency).
$t$ is the time.
$\phi$ is the phase constant.

(A) The instantaneous current in an $AC$ circuit is given by $I(t) = I_{max} \sin(\omega t)$.
To find the sum (or integral) of the instantaneous current values over one complete cycle of period $T$,we calculate the integral $\int_{0}^{T} I(t) dt$.
$\int_{0}^{T} I_{max} \sin(\omega t) dt = I_{max} [-\frac{\cos(\omega t)}{\omega}]_{0}^{T}$.
Since $\omega = \frac{2\pi}{T}$,we have $\omega T = 2\pi$.
Substituting the limits: $I_{max} [-\frac{\cos(2\pi)}{\omega} - (-\frac{\cos(0)}{\omega})] = I_{max} [-\frac{1}{\omega} + \frac{1}{\omega}] = 0$.
Therefore,the sum of the instantaneous current values over one complete $AC$ cycle is $0$.

(N/A) The $rms$ (root mean square) value of an alternating current is defined as the square root of the mean of the squares of the instantaneous currents over one complete cycle.
It is also known as the virtual or effective value of the alternating current.
It represents the value of a steady direct current which,when passed through a resistor for a given time,produces the same amount of heat as the alternating current does in the same resistor and in the same time.
The formula for $rms$ current $(I_{rms})$ in terms of peak current $(I_0)$ is given by:
$I_{rms} = \frac{I_0}{\sqrt{2}} \approx 0.707 I_0$

(B) The given voltage $V_{rms} = 220\, V$ represents the root mean square value of the alternating voltage.
The relationship between the peak voltage $(V_0)$ and the root mean square voltage $(V_{rms})$ is given by the formula: $V_0 = V_{rms} \times \sqrt{2}$.
Substituting the given value: $V_0 = 220 \times 1.414$.
$V_0 \approx 311.08\, V$.
Therefore,the maximum voltage is approximately $311\, V$.

(B) In India,the frequency of $AC$ supply is $f = 50 \ Hz$.
This means the current completes $50$ cycles in $1$ $sec$.
In one complete cycle of a sine wave,the voltage passes through zero twice (once at the start/end and once at the half-cycle point).
Therefore,the number of times the voltage becomes zero in $1$ $sec$ is $2 \times f$.
Calculation: $2 \times 50 = 100$ times.

(N/A) The given waveform is a periodic square wave. The current $I(t)$ takes the value $I_1 = 1 \text{ A}$ for time interval $0 < t < T/2$ and $I_2 = -2 \text{ A}$ for time interval $T/2 < t < T$.
The root mean square current $I_{rms}$ is defined as:
$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2(t) dt}$
Substituting the values:
$I_{rms} = \sqrt{\frac{1}{T} \left[ \int_{0}^{T/2} (1)^2 dt + \int_{T/2}^{T} (-2)^2 dt \right]}$
$I_{rms} = \sqrt{\frac{1}{T} \left[ (1 \times T/2) + (4 \times T/2) \right]}$
$I_{rms} = \sqrt{\frac{1}{T} \left[ \frac{T}{2} + 2T \right]} = \sqrt{\frac{1}{T} \left( \frac{5T}{2} \right)}$
$I_{rms} = \sqrt{2.5} \approx 1.58 \text{ A}$.

(N/A) For a direct current $(DC)$,$1$ ampere is defined as $1$ coulomb of charge flowing per second.
An alternating current $(AC)$ changes its direction periodically with the source frequency. If we were to measure the average current,it would be zero over a full cycle,which is not useful for power measurement.
Therefore,the $AC$ ampere is defined based on a property that is independent of the direction of current,which is the heating effect (Joule's heating).
$1$ ampere of $AC$ is defined as the value of alternating current that produces the same amount of heat in a given resistor as $1$ ampere of $DC$ would produce in the same resistor over the same time interval. This is known as the root-mean-square $(RMS)$ value of the current.

(B) The given current is $I = I_{1} \sin \omega t + I_{2} \cos \omega t$.
This can be rewritten in the form $I = I_{0} \sin(\omega t + \phi)$,where $I_{0}$ is the peak current.
The amplitude $I_{0}$ is given by $\sqrt{I_{1}^{2} + I_{2}^{2}}$.
$A$ hot wire ammeter measures the root mean square $(RMS)$ value of the current.
The $RMS$ value is defined as $I_{rms} = \frac{I_{0}}{\sqrt{2}}$.
Substituting the value of $I_{0}$,we get $I_{rms} = \sqrt{\frac{I_{1}^{2} + I_{2}^{2}}{2}}$.

(A) The instantaneous current in an $AC$ circuit is given by $i = i_{0} \cos(\omega t)$,where $i_{0}$ is the peak current.
At $t = 0$,the current is at its maximum value,$i = i_{0}$.
The $rms$ value of the current is $i_{rms} = \frac{i_{0}}{\sqrt{2}}$.
We need to find the time $t$ when $i = \frac{i_{0}}{\sqrt{2}}$.
Setting $\cos(\omega t) = \frac{1}{\sqrt{2}}$,we get $\omega t = \frac{\pi}{4}$.
Substituting $\omega = 2\pi f$,we have $2\pi f t = \frac{\pi}{4}$.
Solving for $t$,$t = \frac{1}{8f}$.
Given $f = 50\, Hz$,$t = \frac{1}{8 \times 50} = \frac{1}{400}\, s$.
$t = 0.0025\, s = 2.5\, ms$.

(A) The given equation is $i = i_{1} \sin \omega t + i_{2} \cos \omega t$.
We can rewrite $\cos \omega t$ as $\sin(\omega t + 90^{\circ})$.
So,$i = i_{1} \sin \omega t + i_{2} \sin(\omega t + 90^{\circ})$.
This represents the superposition of two sinusoidal currents with a phase difference of $90^{\circ}$.
The resultant peak current $i_{0}$ is given by $i_{0} = \sqrt{i_{1}^{2} + i_{2}^{2} + 2i_{1}i_{2} \cos(90^{\circ})}$.
Since $\cos(90^{\circ}) = 0$,we get $i_{0} = \sqrt{i_{1}^{2} + i_{2}^{2}}$.
The root mean square (rms) current is defined as $i_{rms} = \frac{i_{0}}{\sqrt{2}}$.
Therefore,$i_{rms} = \frac{\sqrt{i_{1}^{2} + i_{2}^{2}}}{\sqrt{2}} = \frac{1}{\sqrt{2}}(i_{1}^{2} + i_{2}^{2})^{1/2}$.

(A) The given current is $i = i_1 + i_2$,where $i_1 = \sqrt{42} \sin \left(\frac{2 \pi}{T} t\right)$ and $i_2 = 10$.
The $r.m.s.$ value of a composite current $i = i_1 + i_2$ is given by $I_{rms} = \sqrt{I_{1,rms}^2 + I_{2,rms}^2}$.
For the sinusoidal component $i_1$,the $r.m.s.$ value is $I_{1,rms} = \frac{I_0}{\sqrt{2}} = \frac{\sqrt{42}}{\sqrt{2}} = \sqrt{21}$.
For the constant component $i_2 = 10$,the $r.m.s.$ value is $I_{2,rms} = 10$.
Therefore,$I_{rms} = \sqrt{(\sqrt{21})^2 + 10^2} = \sqrt{21 + 100} = \sqrt{121} = 11 \text{ A}$.

(A) The instantaneous current in an $AC$ circuit is given by $i = i_0 \sin(\omega t)$.
At maximum value,$i = i_0$,so $i_0 = i_0 \sin(\omega t_1) \Rightarrow \omega t_1 = \frac{\pi}{2}$.
At rms value,$i = \frac{i_0}{\sqrt{2}}$,so $\frac{i_0}{\sqrt{2}} = i_0 \sin(\omega t_2) \Rightarrow \omega t_2 = \frac{\pi}{4}$.
The time taken to change from maximum value to rms value is $\Delta t = t_1 - t_2$.
$\Delta t = \frac{\pi}{2\omega} - \frac{\pi}{4\omega} = \frac{\pi}{4\omega}$.
Since $\omega = 2\pi f$,we have $\Delta t = \frac{\pi}{4(2\pi f)} = \frac{1}{8f}$.
Given $f = 50\, Hz$,$\Delta t = \frac{1}{8 \times 50} = \frac{1}{400}\, s$.
Converting to milliseconds: $\Delta t = \frac{1}{400} \times 1000\, ms = 2.5\, ms$.