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

(B) In a $DC$ ammeter,a coil is free to rotate in the magnetic field of a fixed magnet.
If an alternating current is passed through such a coil,the torque will reverse its direction each time the current changes direction.
Since the frequency of $AC$ is typically high,the coil cannot follow the rapid changes in torque due to its inertia.
Consequently,the average value of the torque over a complete cycle is zero,and the pointer remains at the zero position.

(B) The instantaneous current is given by the equation $i = i_0 \cos(\omega t + \phi)$,where $i_0$ is the peak current.
Comparing this with the given equation $i = 4 \cos(\omega t + \phi)$,we find the peak current $i_0 = 4 \text{ A}$.
The $r.m.s.$ value of an alternating current is related to the peak current by the formula $i_{r.m.s.} = \frac{i_0}{\sqrt{2}}$.
Substituting the value of $i_0$,we get $i_{r.m.s.} = \frac{4}{\sqrt{2}} = \frac{4 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \text{ A}$.
Therefore,the correct option is $B$.

(C) The effective voltage (also known as the root mean square voltage,$V_{rms}$) is related to the peak voltage $(V_0)$ by the formula: $V_{rms} = \frac{V_0}{\sqrt{2}}$.
Given that the peak voltage $V_0 = 423 \ V$.
Substituting the value into the formula:
$V_{rms} = \frac{423}{\sqrt{2}} \approx \frac{423}{1.414} \approx 299.15 \ V$.
Rounding to the nearest whole number,we get $V_{rms} = 300 \ V$.
Therefore,the correct option is $C$.

(C) The relationship between the peak current $(I_0)$ and the root mean square current $(I_{rms})$ for a sinusoidal alternating current is given by the formula:
$I_{rms} = \frac{I_0}{\sqrt{2}}$
Given that the peak value $I_0 = 6 \ A$,we substitute this value into the formula:
$I_{rms} = \frac{6}{\sqrt{2}}$
To simplify,multiply the numerator and denominator by $\sqrt{2}$:
$I_{rms} = \frac{6 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \ A$
Therefore,the r.m.s. value of the current is $3\sqrt{2} \ A$.

(C) The given voltage equation is $V = 240 \sin(120t)$.
Comparing this with the standard form $V = V_m \sin(\omega t)$,we get the peak voltage $V_m = 240 \text{ V}$ and angular frequency $\omega = 120 \text{ rad/s}$.
The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{120}{2 \times 3.14159} \approx 19.1 \text{ Hz}$.
The $r.m.s.$ voltage is given by $V_{rms} = \frac{V_m}{\sqrt{2}} = \frac{240}{1.414} \approx 169.7 \text{ V} \approx 170 \text{ V}$.
Thus,the frequency is $19 \text{ Hz}$ and the $r.m.s.$ voltage is $170 \text{ V}$.

(D) In an alternating current $(AC)$ circuit,the instantaneous voltage is given by $E = E_0 \sin(\omega t)$.
The root mean square $(r.m.s.)$ value of an alternating voltage is defined as the square root of the mean of the squares of the instantaneous values over one complete cycle.
Mathematically,$E_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} E^2 dt}$.
Substituting $E = E_0 \sin(\omega t)$,we get $E_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} E_0^2 \sin^2(\omega t) dt}$.
Solving this integral,we obtain $E_{rms} = \frac{E_0}{\sqrt{2}}$.

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

(C) The root mean square $(RMS)$ current is given by Ohm's law: $I_{rms} = \frac{V_{rms}}{R} = \frac{200 \ V}{40 \ \Omega} = 5 \ A$.
The peak current $(I_0)$ is related to the $RMS$ current by the formula: $I_0 = I_{rms} \times \sqrt{2}$.
Substituting the values: $I_0 = 5 \times 1.414 = 7.07 \ A$.
Therefore,the peak value of the electric current is approximately $7 \ A$.

(B) The standard frequency of alternating current $(AC)$ supplied in the power mains in India is $50 \text{ Hz}$. This means the current completes $50$ cycles per second.

(D) The time period $T$ of the alternating current is given by $T = \frac{1}{\nu} = \frac{1}{50} \ s = 0.02 \ s$.
The time taken to reach the maximum value from zero is one-fourth of the time period:
$t = \frac{T}{4} = \frac{0.02}{4} = 0.005 \ s = 5 \times 10^{-3} \ s$.
The peak value of the current $(i_0)$ is related to the $r.m.s.$ value $(i_{rms})$ by the formula:
$i_0 = i_{rms} \times \sqrt{2} = 10 \times 1.414 = 14.14 \ A$.
Therefore,the time taken is $5 \times 10^{-3} \ s$ and the peak value is $14.14 \ A$.

(C) The root mean square $(RMS)$ value of an alternating current is defined as the value of the steady current which,when flowing through a given resistor for a given time,produces the same amount of heat as the alternating current does over the same period.
Mathematically,for a sinusoidal alternating current $I = I_0 \sin(\omega t)$,the $RMS$ value $(I_{rms})$ is given by:
$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2 dt}$
Solving this integral over one complete cycle yields:
$I_{rms} = \frac{I_0}{\sqrt{2}}$
Where $I_0$ is the peak value of the current.
Therefore,the $RMS$ value is $\frac{1}{\sqrt{2}}$ times the peak value.

(B) Given: Peak value $E_0 = 10 \ V$,frequency $f = 50 \ Hz$,and time $t = \frac{1}{600} \ s$.
The instantaneous e.m.f. is given by $E = E_0 \cos(\omega t)$.
Since $\omega = 2\pi f$,we have $\omega = 2 \times \pi \times 50 = 100\pi \ rad/s$.
Substituting the values into the equation:
$E = 10 \cos(100\pi \times \frac{1}{600})$
$E = 10 \cos(\frac{\pi}{6})$
Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we get:
$E = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \ V$.

(C) The standard equation for an $ac$ voltage is given by $E = E_0 \sin(\omega t)$,where $\omega$ is the angular frequency.
Comparing this with the given equation $E = 170 \sin(377t)$,we get $\omega = 377 \text{ rad/s}$.
The relationship between angular frequency $\omega$ and frequency $\nu$ is $\omega = 2\pi \nu$.
Therefore,$\nu = \frac{\omega}{2\pi} = \frac{377}{2 \times 3.14159} \approx 60.03 \text{ Hz}$.
Rounding to the nearest integer,the frequency is $60 \text{ Hz}$.

(A) For an alternating current given by $I = I_0 \sin(\omega t)$,the average value of current over a complete cycle is given by $I_{avg} = \frac{1}{T} \int_0^T I_0 \sin(\omega t) dt = 0$. This is because the positive and negative half-cycles cancel each other out.
The average value of the square of the current is $I_{rms}^2 = \frac{I_0^2}{2}$,which is not zero.
Average power dissipation is given by $P_{avg} = V_{rms} I_{rms} \cos(\phi)$,which is generally not zero unless the circuit is purely reactive.
The phase difference between voltage and current is not necessarily zero; it depends on the components $(R, L, C)$ in the circuit.

(C) The given equation is $i = i_1 \cos \omega t + i_2 \sin \omega t$.
This can be rewritten as $i = I_0 \sin(\omega t + \phi)$,where the amplitude $I_0 = \sqrt{i_1^2 + i_2^2}$.
The root mean square (r.m.s.) current is defined as $i_{rms} = \frac{I_0}{\sqrt{2}}$.
Substituting the value of $I_0$,we get $i_{rms} = \frac{\sqrt{i_1^2 + i_2^2}}{\sqrt{2}} = \frac{1}{\sqrt{2}}(i_1^2 + i_2^2)^{1/2}$.

(C) The root mean square voltage is given as $V_{rms} = 220 V$.
The relationship between the peak voltage $(V_0)$ and the $RMS$ voltage $(V_{rms})$ is $V_0 = V_{rms} \times \sqrt{2}$.
Substituting the values: $V_0 = 220 \times 1.414$.
$V_0 = 311.08 V$.
Rounding to the nearest integer,the peak voltage is $311 V$.
Therefore,the correct option is $C$.

(C) hot wire ammeter measures the root mean square $(rms)$ value of an alternating current.
Given,$I_{rms} = 10 \ A$.
The relationship between the peak value $(I_0)$ and the $rms$ value $(I_{rms})$ is given by $I_0 = I_{rms} \times \sqrt{2}$.
Substituting the value: $I_0 = 10 \times 1.414 = 14.14 \ A$.
Therefore,the peak value of the current is $14.14 \ A$.

(D) In an alternating current $(AC)$ circuit,the voltage provided for domestic use is specified as $220\,V$.
This value represents the Root Mean Square $(RMS)$ voltage of the supply.
The $RMS$ value is defined as the square root of the mean of the squares of the instantaneous voltages over one complete cycle.
It is the equivalent $DC$ voltage that would produce the same amount of heat in a resistor as the $AC$ voltage.

(B) The root mean square (r.m.s.) voltage is given as $V_{rms} = 220 \ V$.
The relationship between the peak voltage $(V_0)$ and the r.m.s. voltage is $V_0 = \sqrt{2} \times V_{rms}$.
Substituting the given value: $V_0 = 1.414 \times 220 \ V$.
$V_0 \approx 311 \ V$.
Therefore,electrical appliances should be designed to withstand an instantaneous peak voltage of approximately $311 \ V$,which is closest to $310 \ V$ among the given options.

(B) The $r.m.s.$ value of an alternating voltage is related to its peak amplitude $(V_0)$ by the formula: $V_{rms} = \frac{V_0}{\sqrt{2}}$.
Given the amplitude $V_0 = 120 \text{ V}$.
Substituting the value: $V_{rms} = \frac{120}{1.414} \approx 84.8 \text{ V}$.
Thus,the correct option is $B$.

(D) The peak value of current is $i_0$ and the r.m.s. value is $i_{rms} = \frac{i_0}{\sqrt{2}}$.
Given the alternating current equation $i = i_0 \sin(100 \pi t)$.
At peak value,the phase $\omega t_1 = \frac{\pi}{2}$,so $t_1 = \frac{\pi}{2 \times 100 \pi} = \frac{1}{200} \, \text{s}$.
At r.m.s. value,$i = \frac{i_0}{\sqrt{2}}$,so $\sin(100 \pi t_2) = \frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4})$.
Thus,$100 \pi t_2 = \frac{\pi}{4} \Rightarrow t_2 = \frac{1}{400} \, \text{s}$.
The time taken is $\Delta t = t_1 - t_2 = \frac{1}{200} - \frac{1}{400} = \frac{2-1}{400} = \frac{1}{400} \, \text{s}$.
$\Delta t = 0.0025 \, \text{s} = 2.5 \times 10^{-3} \, \text{s}$.

(C) The given equations are $V = 5\sin(100\pi t - \frac{\pi}{6})$ and $I = 4\sin(100\pi t + \frac{\pi}{6})$.
Comparing these with the standard forms $V = V_m\sin(\omega t + \phi_1)$ and $I = I_m\sin(\omega t + \phi_2)$,we get the phase angles as $\phi_1 = -\frac{\pi}{6}$ and $\phi_2 = \frac{\pi}{6}$.
The phase difference is $\Delta \phi = \phi_2 - \phi_1 = \frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.
Since $\frac{\pi}{3}$ radians is equal to $60^o$ and $\phi_2 > \phi_1$,the current leads the voltage by $60^o$.

(A) The given $AC$ supply voltage is the root mean square value,$V_{rms} = 220 \ V$.
The peak voltage $V_0$ is given by $V_0 = V_{rms} \times \sqrt{2}$.
The average e.m.f. $(V_{av})$ over a positive half cycle is given by the formula $V_{av} = \frac{2}{\pi} V_0$.
Substituting the value of $V_0$,we get $V_{av} = \frac{2}{\pi} \times (V_{rms} \times \sqrt{2}) = \frac{2\sqrt{2}}{\pi} V_{rms}$.
Substituting $V_{rms} = 220 \ V$,we get $V_{av} = \frac{2 \times 1.414}{3.1416} \times 220 \approx 198 \ V$.

(B) The $r.m.s.$ (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.
For a sinusoidal alternating current given by $I = I_0 \sin(\omega t)$,the $r.m.s.$ value is calculated as:
$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2 dt} = \sqrt{\frac{1}{T} \int_{0}^{T} I_0^2 \sin^2(\omega t) dt}$
Solving this integral over one period $T$ gives:
$I_{rms} = \frac{I_0}{\sqrt{2}}$
Therefore,the correct relation is $I_{rms} = \frac{1}{\sqrt{2}} I_0$.

(A) The given alternating voltage is $E = 20 \sin 300t$.
The average value of a sinusoidal alternating voltage over one complete cycle is defined as the integral of the voltage over the time period $T$ divided by the time period $T$.
Mathematically,$E_{avg} = \frac{1}{T} \int_{0}^{T} E_0 \sin(\omega t) dt$.
Since the sine function is symmetric about the time axis,the area under the positive half-cycle is equal to the area under the negative half-cycle.
Therefore,the positive and negative values cancel each other out,resulting in an average value of $0$ over one complete cycle.

(C) For an alternating current,the relationship between the peak value $(I_0)$ and the root mean square value $(I_{rms})$ is given by $I_{rms} = \frac{I_0}{\sqrt{2}}$.
Rearranging this formula to find the ratio of the peak value to the r.m.s. value:
$\frac{I_0}{I_{rms}} = \sqrt{2}$.
Therefore,the ratio is $\sqrt{2}$.

(A) The given voltage $V_{rms} = 200 \ V$ and resistance $R = 280 \ \Omega$.
First,calculate the root mean square current $(I_{rms})$:
$I_{rms} = \frac{V_{rms}}{R} = \frac{200}{280} = \frac{5}{7} \ A \approx 0.714 \ A$.
The peak value of current $(I_0)$ is related to $I_{rms}$ by the formula $I_0 = I_{rms} \times \sqrt{2}$.
$I_0 = \frac{5}{7} \times 1.414 \approx 0.714 \times 1.414 \approx 1.01 \ A$.
Therefore,the peak value of current is about $1 \ A$.

(D) The frequency of the $ac$ source is $f = 50 \text{ Hz}$.
The time period $T$ of the $ac$ cycle is given by $T = 1/f = 1/50 \text{ s} = 0.02 \text{ s}$.
The voltage varies as $V = V_0 \sin(\omega t)$. The peak value occurs at $t = T/4$ and the voltage is zero at $t = T/2$ (or at $t=0$ for the start of the cycle).
The time taken to go from the peak value $(V = V_0)$ to zero $(V = 0)$ corresponds to a quarter of the time period.
Therefore,the required time $t = T/4 = 0.02 / 4 = 0.005 \text{ s} = 5 \times 10^{-3} \text{ s}$.

(B) In an $AC$ circuit,the given value of potential is considered to be the root mean square $(RMS)$ value unless specified otherwise.
Given: $V_{rms} = 10 \ V$.
The relationship between the peak value $(V_0)$ and the $RMS$ value $(V_{rms})$ is given by the formula: $V_0 = \sqrt{2} \times V_{rms}$.
Substituting the given value: $V_0 = \sqrt{2} \times 10 = 10\sqrt{2} \ V$.
Therefore,the peak value of the potential is $10\sqrt{2} \ V$.

(C) The given equation for alternating voltage is $E = 141 \sin(628 t)$.
Comparing this with the standard form $E = E_0 \sin(\omega t)$,we get the peak voltage $E_0 = 141 \text{ V}$ and angular frequency $\omega = 628 \text{ rad/s}$.
The root mean square (rms) voltage is given by $E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{141}{1.414} \approx 100 \text{ V}$.
The angular frequency is related to frequency $f$ by the formula $\omega = 2 \pi f$.
Substituting the values,$628 = 2 \times 3.14 \times f$.
$f = \frac{628}{6.28} = 100 \text{ Hz}$.
Thus,the rms voltage is $100 \text{ V}$ and the frequency is $100 \text{ Hz}$.

(C) The relationship between the peak voltage $(E_0)$ and the root mean square voltage $(E_{rms})$ is given by the formula: $E_{rms} = \frac{E_0}{\sqrt{2}}$.
Given that the maximum voltage $E_0 = 707 \ V$.
Substituting the value: $E_{rms} = \frac{707}{1.414} \approx 500 \ V$.
Therefore,the correct option is $C$.

(D) Given the voltage equation: $V = 120 \sin(100\pi t) \cos(100\pi t)$.
Using the trigonometric identity $\sin(2\theta) = 2 \sin \theta \cos \theta$,we can rewrite the equation as:
$V = 60 \times (2 \sin(100\pi t) \cos(100\pi t))$
$V = 60 \sin(200\pi t)$.
Comparing this with the standard form $V = V_{\max} \sin(\omega t)$,we get:
Maximum voltage $V_{\max} = 60 \, V$.
Angular frequency $\omega = 200\pi \, rad/s$.
Since $\omega = 2\pi \nu$,the frequency $\nu$ is:
$\nu = \frac{\omega}{2\pi} = \frac{200\pi}{2\pi} = 100 \, Hz$.
Thus,the maximum voltage is $60 \, V$ and the frequency is $100 \, Hz$.

(C) The r.m.s. value of current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_{t_1}^{t_2} i^2 \ dt}$,where $T = t_2 - t_1 = 4 - 2 = 2 \ s$.
First,calculate the mean square current $\overline{i^2}$:
$\overline{i^2} = \frac{1}{2} \int_{2}^{4} (2\sqrt{t})^2 \ dt = \frac{1}{2} \int_{2}^{4} 4t \ dt$
$\overline{i^2} = 2 \int_{2}^{4} t \ dt = 2 \left[ \frac{t^2}{2} \right]_{2}^{4} = [t^2]_{2}^{4} = 4^2 - 2^2 = 16 - 4 = 12 \ A^2$.
Now,find the r.m.s. value:
$i_{rms} = \sqrt{\overline{i^2}} = \sqrt{12} = 2\sqrt{3} \ A$.

(B) The $r.m.s.$ value of a current $I(t)$ is given by $I_{rms} = \sqrt{\frac{1}{T} \int_0^T I^2(t) dt}$.
For $(1)$,$I = x_0 \sin \omega t$. The $r.m.s.$ value is $\frac{x_0}{\sqrt{2}}$. Thus,$1-(ii)$.
For $(2)$,$I = x_0 \sin \omega t \cos \omega t = \frac{x_0}{2} \sin(2\omega t)$. The peak value is $\frac{x_0}{2}$,so the $r.m.s.$ value is $\frac{x_0/2}{\sqrt{2}} = \frac{x_0}{2\sqrt{2}}$. Thus,$2-(iii)$.
For $(3)$,$I = x_0 \sin \omega t + x_0 \cos \omega t = \sqrt{2} x_0 \sin(\omega t + \frac{\pi}{4})$. The peak value is $\sqrt{2} x_0$,so the $r.m.s.$ value is $\frac{\sqrt{2} x_0}{\sqrt{2}} = x_0$. Thus,$3-(i)$.
Therefore,the correct match is $1-(ii), 2-(iii), 3-(i)$.

(C) The $r.m.s.$ value of a current $i(t)$ over a time interval $T$ is given by the formula: $i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} i^2 dt}$.
Given $i = t^2$,we substitute this into the formula:
$i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} (t^2)^2 dt} = \sqrt{\frac{1}{T} \int_{0}^{T} t^4 dt}$.
Performing the integration:
$\int_{0}^{T} t^4 dt = \left[ \frac{t^5}{5} \right]_{0}^{T} = \frac{T^5}{5}$.
Now,substitute this back into the $r.m.s.$ expression:
$i_{rms} = \sqrt{\frac{1}{T} \cdot \frac{T^5}{5}} = \sqrt{\frac{T^4}{5}} = \frac{T^2}{\sqrt{5}}$.
Thus,the correct option is $C$.

(B) From the given figure,the emf $(E)$ starts from zero at $\omega t = 0$ and reaches its maximum at $\omega t = \pi / 2$. Thus,$E = E_0 \sin(\omega t)$.
The current $(I)$ starts from its minimum value at $\omega t = 0$ and reaches zero at $\omega t = \pi / 2$. This indicates that the current follows the equation $I = I_0 \sin(\omega t - \pi / 2)$.
Comparing the phases,the phase of the voltage is $\omega t$ and the phase of the current is $(\omega t - \pi / 2)$.
Therefore,the voltage leads the current by a phase angle of $\pi / 2$.

(A) The given waveform is a square wave with amplitude $V_0 = 10 \ V$.
For a square wave that oscillates between $+V_0$ and $-V_0$,the instantaneous voltage $V(t)$ is either $+V_0$ or $-V_0$.
The $r.m.s.$ voltage is defined as $V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2(t) dt}$.
Since $V^2(t) = V_0^2$ at all times,
$V_{rms} = \sqrt{\frac{1}{T} \int_0^T V_0^2 dt} = \sqrt{\frac{V_0^2}{T} \cdot T} = V_0$.
Given $V_0 = 10 \ V$,therefore $V_{rms} = 10 \ V$.

(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 $\nu = \frac{1}{T} = \frac{1}{0.4} = 2.5 \, Hz$.
Wave $M$ starts at $t = 0$ with zero value and positive slope, representing $\sin(\omega t)$.
Wave $N$ starts at $t = 0$ with a negative value, reaching its peak later than $M$. Specifically, $N$ is delayed by a quarter of a cycle compared to $M$.
A delay of a quarter cycle corresponds to a phase lag of $\frac{\pi}{2}$ radians.
Thus, the phase lead of $N$ over $M$ is $-\frac{\pi}{2}$ radians.

(C) The heat produced by an $AC$ current in a resistor $R$ over time $t$ is given by $H_{AC} = I_{rms}^2 Rt$.
The heat produced by a $DC$ current $I_{DC} = 2 \, A$ in the same resistor $R$ over the same time $t$ is $H_{DC} = I_{DC}^2 Rt$.
According to the problem,$H_{AC} = 3 \times H_{DC}$.
Substituting the formulas: $I_{rms}^2 Rt = 3 \times (2^2) Rt$.
Canceling $R$ and $t$ from both sides: $I_{rms}^2 = 3 \times 4 = 12$.
Therefore,$I_{rms} = \sqrt{12} = 2\sqrt{3} \approx 3.46 \, A$.

(D) The given equation is $I = I_0 \sin(\omega t)$,where $I_0 = 100 \ A$ and $\omega = 200 \pi \ rad/s$.
The current reaches its maximum value when $\sin(\omega t) = 1$,which occurs at $\omega t = \frac{\pi}{2}$.
Substituting the value of $\omega$: $(200 \pi) t = \frac{\pi}{2}$.
Solving for $t$: $t = \frac{\pi}{2 \times 200 \pi} = \frac{1}{400} \ s$.

(D) The time period $T$ of the $AC$ source is given by $T = 1/f = 1/50 = 0.02\,sec$.
The voltage reaches its maximum value at $t = T/4$ and reaches zero at $t = T/2$ (relative to the start of the cycle).
However,the time taken to go from the maximum value to zero is exactly one-quarter of the time period.
Time $t = T/4 = 0.02 / 4 = 0.005\,sec = 5 \times 10^{-3}\,sec$.

(C) The $r.m.s.$ value of current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_{t_1}^{t_2} i^2 \ dt}$,where $T = t_2 - t_1 = 4 - 2 = 2 \ s$.
First,calculate the mean square value $\overline{i^2} = \frac{1}{T} \int_{2}^{4} (2\sqrt{t})^2 \ dt$.
$\overline{i^2} = \frac{1}{2} \int_{2}^{4} 4t \ dt = 2 \int_{2}^{4} t \ dt$.
Evaluating the integral: $2 \left[ \frac{t^2}{2} \right]_{2}^{4} = [t^2]_{2}^{4} = 4^2 - 2^2 = 16 - 4 = 12$.
Finally,$i_{rms} = \sqrt{\overline{i^2}} = \sqrt{12} = 2\sqrt{3} \ A$.

(C) The $r.m.s.$ value of a current $i(t)$ over a time interval $T$ is given by the formula: $i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} i^2 dt}$.
Given $i = t^2$,we have $i^2 = t^4$.
Substituting this into the formula:
$i_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} t^4 dt}$.
Evaluating the integral: $\int_{0}^{T} t^4 dt = \left[ \frac{t^5}{5} \right]_{0}^{T} = \frac{T^5}{5}$.
Now,substitute back into the expression:
$i_{rms} = \sqrt{\frac{1}{T} \cdot \frac{T^5}{5}} = \sqrt{\frac{T^4}{5}} = \frac{T^2}{\sqrt{5}}$.

(C) Given the potential difference $V$ as a function of time $t$:
$V = V_0$ for $0 \leq t \leq \frac{T}{2}$
$V = 0$ for $\frac{T}{2} \leq t \leq T$
The $r.m.s.$ value is defined as:
$V_{rms} = \sqrt{\frac{1}{T} \int_0^T V^2 dt}$
Substituting the given values:
$V_{rms} = \sqrt{\frac{1}{T} \left( \int_0^{T/2} V_0^2 dt + \int_{T/2}^T 0^2 dt \right)}$
$V_{rms} = \sqrt{\frac{1}{T} \left( V_0^2 [t]_0^{T/2} + 0 \right)}$
$V_{rms} = \sqrt{\frac{V_0^2}{T} \cdot \frac{T}{2}}$
$V_{rms} = \sqrt{\frac{V_0^2}{2}}$
$V_{rms} = \frac{V_0}{\sqrt{2}}$

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

(A) The instantaneous current is given by $I = I_0 \sin(\omega t + \phi_I)$,where $\phi_I = 0$.
The instantaneous emf is given by $E = E_0 \sin(\omega t + \phi_E)$,where $\phi_E = -\pi/6$.
The phase difference between $E$ and $I$ is defined as $\Delta \phi = \phi_E - \phi_I$.
Substituting the values,we get $\Delta \phi = (-\pi/6) - (0) = -\pi/6 \, rad$.
Therefore,the phase difference is $-\pi/6 \, rad$.

(C) The heat produced by an alternating current $(ac)$ is given by $H_{ac} = i_{rms}^2Rt$.
The heat produced by a direct current $(dc)$ is given by $H_{dc} = i^2Rt$.
According to the problem,$H_{ac} = 3 \times H_{dc}$.
Substituting the formulas: $i_{rms}^2Rt = 3 \times i^2Rt$.
Canceling $R$ and $t$ from both sides,we get $i_{rms}^2 = 3 \times i^2$.
Given $i = 2 \ A$,we have $i_{rms}^2 = 3 \times (2)^2 = 3 \times 4 = 12$.
Therefore,$i_{rms} = \sqrt{12} = 2\sqrt{3} \approx 3.46 \ A$.

(D) An $AC$ ammeter is designed to measure the root-mean-square $(RMS)$ value of an alternating current.
In contrast,a $DC$ ammeter is designed to measure the average value of current over a complete cycle.
For a sinusoidal alternating current,the average value over one complete cycle is defined as $I_{avg} = \frac{1}{T} \int_{0}^{T} I_{0} \sin(\omega t) dt = 0$.
Since the $DC$ ammeter measures the average current,it will show a reading of zero.

(A) The $rms$ value of a current $I(t)$ over a time period $T$ is defined as $I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I(t)^2 dt}$.
Given $I = I_0 + I_1 \sin \omega t$, we square the expression: $I^2 = I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t$.
Now, integrate over one full cycle $T = \frac{2\pi}{\omega}$:
$I_{rms}^2 = \frac{1}{T} \int_{0}^{T} (I_0^2 + I_1^2 \sin^2 \omega t + 2 I_0 I_1 \sin \omega t) dt$.
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, $I_{rms}^2 = I_0^2 + I_1^2 \left( \frac{1}{2} \right) = I_0^2 + 0.5 I_1^2$.
Therefore, $I_{rms} = \sqrt{I_0^2 + 0.5 I_1^2}$.

(C) The $r.m.s.$ (root mean square) value of a current is defined as the value of the steady direct current which,when flowing through a given resistor for a given time,produces the same amount of heat as the alternating current does in the same resistor for the same time.
For the given square wave current:
$I(t) = 2 \text{ A}$ for $0 < t < 0.01 \text{ s}$
$I(t) = -2 \text{ A}$ for $0.01 < t < 0.02 \text{ s}$
The time period $T = 0.02 \text{ s}$.
The $r.m.s.$ current is given by the formula:
$I_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} I^2(t) dt}$
$I_{rms} = \sqrt{\frac{1}{0.02} \left[ \int_{0}^{0.01} (2)^2 dt + \int_{0.01}^{0.02} (-2)^2 dt \right]}$
$I_{rms} = \sqrt{\frac{1}{0.02} \left[ 4 \times 0.01 + 4 \times 0.01 \right]}$
$I_{rms} = \sqrt{\frac{1}{0.02} \left[ 0.04 + 0.04 \right]}$
$I_{rms} = \sqrt{\frac{0.08}{0.02}} = \sqrt{4} = 2 \text{ A}$.
Thus,the $r.m.s.$ current is $2 \text{ A}$.