Application of junction diode (Rectifier) Questions in English

(C) The process of converting alternating current $(ac)$ into direct current $(dc)$ is called $Rectification$. This process is typically performed using electronic components such as diodes,which allow current to flow in only one direction.

(B) rectifier circuit converts $ac$ into $dc$,but the output is pulsating in nature. To obtain a smooth $dc$ output,a filter circuit is used. $A$ filter circuit removes the $ac$ ripples from the rectified output. Common examples include capacitor filters,inductor filters,and $\pi$-filters.

(B) $PN$ junction diode allows current to flow in only one direction (when forward-biased) and blocks it in the opposite direction (when reverse-biased). Due to this unidirectional property,it is primarily used as a rectifier to convert alternating current $(AC)$ into direct current $(DC)$.

(D) The efficiency of a half-wave rectifier is $40.6\%$,whereas the efficiency of a full-wave rectifier is $81.2\%$.
The average $DC$ output voltage for a half-wave rectifier is $V_{dc} = V_m / \pi$,while for a full-wave rectifier it is $V_{dc} = 2V_m / \pi$.
Since the full-wave rectifier has higher efficiency and a higher average output voltage compared to the half-wave rectifier,none of the given options are correct.

(A) rectifier is an electrical device that converts alternating current $(ac)$,which periodically reverses direction,into direct current $(dc)$,which flows in only one direction. This process is known as rectification. Therefore,the correct function is to convert $ac$ into $dc$.

(B) half-wave rectifier uses a single diode to convert alternating current $(AC)$ into pulsating direct current $(DC)$.
In a half-wave rectifier circuit,the diode conducts only during the positive half-cycle of the input $AC$ signal and blocks the current during the negative half-cycle.
Diagram $B$ shows a transformer connected to a single diode in series with the load,which is the standard configuration for a half-wave rectifier.
Therefore,the correct diagram is $B$.

(D) The efficiency $\eta$ of a full-wave rectifier is given by the formula: $\eta = \frac{81.2}{1 + \frac{r_f}{R_L}} \%$.
Here,$r_f$ is the forward resistance of the diode and $R_L$ is the load resistance.
For an ideal diode,the forward resistance $r_f$ is negligible compared to the load resistance $R_L$ (i.e.,$r_f \ll R_L$).
Therefore,the maximum efficiency is $\eta_{\max} = 81.2\%$.

(C) In a full wave rectifier,the output consists of two pulses for every single cycle of the input $AC$ supply.
Since the input frequency is $f_{in} = 50\, Hz$,the output ripple frequency $f_{out}$ is given by the formula $f_{out} = 2 \times f_{in}$.
Substituting the given value: $f_{out} = 2 \times 50\, Hz = 100\, Hz$.
Therefore,the fundamental frequency in the ripple is $100\, Hz$.

(C) In a full-wave rectifier,the circuit utilizes both halves of the input $AC$ cycle.
For every single cycle of the input $AC$ signal,the rectifier produces two pulses in the output.
Therefore,the frequency of the output signal is twice the frequency of the input signal.
Output frequency $= 2 \times$ Input frequency.
Given input frequency $= v$.
Thus,the output frequency $= 2v$.

(C) diode is a semiconductor device that allows current to flow in only one direction. Due to this unidirectional property,it is primarily used as a rectifier to convert alternating current $(AC)$ into direct current $(DC)$.

(A) rectifier circuit converts $a.c.$ into a pulsating $d.c.$ signal.
To obtain a smooth $d.c.$ output,we use a filter circuit.
$A$ filter circuit typically consists of components like capacitors and inductors that remove the ripples (fluctuations) from the pulsating $d.c.$ output of the rectifier.
Therefore,the correct answer is $A$ (Filter).

(A) The ripple factor $r$ is defined as the ratio of the root mean square value of the ripple component to the direct current component of the output.
For a half-wave rectifier,the $rms$ value of the output current is $I_{rms} = I_0 / 2$ and the $dc$ value is $I_{dc} = I_0 / \pi$.
The formula for the ripple factor is $r = \sqrt{\left( \frac{I_{rms}}{I_{dc}} \right)^2 - 1}$.
Substituting the values: $r = \sqrt{\left( \frac{I_0 / 2}{I_0 / \pi} \right)^2 - 1} = \sqrt{\left( \frac{\pi}{2} \right)^2 - 1}$.
Since $\pi \approx 3.14$,$\pi / 2 \approx 1.57$.
$r = \sqrt{(1.57)^2 - 1} = \sqrt{2.46 - 1} = \sqrt{1.46} \approx 1.21$.

(A) full wave rectifier uses two diodes to convert both halves of the $AC$ input cycle into a unidirectional $DC$ output.
In a center-tapped full wave rectifier,the two diodes conduct during alternate half-cycles of the input $AC$ signal.
During the first half-cycle,one diode is forward-biased and conducts,while the other is reverse-biased.
During the second half-cycle,the roles are reversed.
Therefore,the two diodes work alternately to rectify the complete cycle.

(B) For a half-wave rectifier,the output voltage is given by $V(t) = V_0 \sin(\omega t)$ for $0 \le t \le T/2$ and $V(t) = 0$ for $T/2 < t < T$.
The $dc$ component (average value) of the output voltage is calculated as:
$V_{dc} = \frac{1}{T} \int_{0}^{T} V(t) dt = \frac{1}{T} \int_{0}^{T/2} V_0 \sin(\omega t) dt$
Since $\omega = 2\pi/T$,we have:
$V_{dc} = \frac{V_0}{T} \left[ -\frac{\cos(\omega t)}{\omega} \right]_{0}^{T/2} = \frac{V_0}{T} \left( -\frac{1}{\omega} (\cos(\pi) - \cos(0)) \right) = \frac{V_0}{T} \left( \frac{2}{\omega} \right) = \frac{V_0}{T} \left( \frac{2}{2\pi/T} \right) = \frac{V_0}{\pi}$.
Given the peak voltage $V_0 = 10 \ V$,the $dc$ component is $V_{dc} = 10/\pi \ V$.

(C) The diode $D$ conducts during the positive half-cycle of the $ac$ supply because it is forward-biased. During the negative half-cycle,it is reverse-biased and does not conduct. The capacitor charges to the peak value of the input voltage during the positive half-cycle. The peak voltage $V_0$ is given by $V_0 = V_{rms} \times \sqrt{2} = 220\sqrt{2} \, V$. Since there is no load resistor to discharge the capacitor,it remains charged to the peak voltage $V_0 = 220\sqrt{2} \, V$. However,if we consider the average $DC$ output voltage for a half-wave rectifier,$V_{dc} = \frac{V_0}{\pi} = \frac{220\sqrt{2}}{\pi} \approx 99 \, V$. Given the options provided,the question likely refers to the peak voltage or a specific $RMS$ interpretation. Re-evaluating the standard interpretation for this specific textbook problem,the capacitor charges to the peak voltage $V_0 = 220\sqrt{2} \approx 311.1 \, V$. Thus,option $(C)$ is the correct answer.

(B) The circuit shown is a half-wave rectifier circuit with a capacitor filter.
In a half-wave rectifier,the capacitor charges to the peak value of the input alternating voltage.
The relationship between the peak voltage $(V_{peak})$ and the root-mean-square voltage $(V_{rms})$ is given by:
$V_{peak} = \sqrt{2} \times V_{rms}$
Given that $V_{rms} = 141.4\,V$,we calculate the peak voltage as:
$V_{peak} = \sqrt{2} \times 141.4$
Since $\sqrt{2} \approx 1.414$,we have:
$V_{peak} = 1.414 \times 141.4 = 200\,V$
Therefore,the maximum potential difference across the condenser is $200\,V$.

(C) The circuit acts as a half-wave rectifier.
When the input voltage is $+10 V$,the $P$-terminal of the diode is at a higher potential than the $N$-terminal,making the diode forward-biased. In this state,the diode conducts,and the output voltage across $R_L$ is $+10 V$.
When the input voltage is $-10 V$,the $P$-terminal is at a lower potential than the $N$-terminal,making the diode reverse-biased. In this state,the diode does not conduct,and the output voltage across $R_L$ is $0 V$.
Therefore,the output signal is a pulse that remains at $+10 V$ when the input is $+10 V$ and drops to $0 V$ when the input is $-10 V$.

(B) half-wave rectifier uses a single diode to convert alternating current $(AC)$ into pulsating direct current $(DC)$.
It allows current to flow only during the positive half-cycle of the input $AC$ signal when the diode is forward-biased.
During the negative half-cycle,the diode is reverse-biased and blocks the current,resulting in zero output voltage across the load resistor $R_L$.
Therefore,the output waveform $V_{CD}$ consists of only the positive half-cycles of the input signal,with zero voltage during the negative half-cycles,which corresponds to the second waveform option.

(D) The circuit acts as a half-wave rectifier with a filter capacitor.
During the positive half-cycle of the input,the diode is forward-biased,and the capacitor charges to the peak voltage of $5 \, V$.
During the negative half-cycle,the diode is reverse-biased and does not conduct.
The capacitor then discharges through the resistor $R$.
The time constant of the circuit is $\tau = RC = (1 \times 10^3 \, \Omega) \times (10 \times 10^{-6} \, F) = 0.01 \, s$.
Assuming the input frequency is standard (e.g.,$50 \, Hz$,period $T = 0.02 \, s$),the time constant $\tau$ is comparable to the period. However,in typical idealized problems of this type,the capacitor is assumed to hold its charge,resulting in a nearly constant $DC$ output voltage of $5 \, V$.

(B) In a full-wave rectifier circuit,during the positive half-cycle of the input $AC$ signal,diode $1$ is forward-biased and conducts,while diode $2$ is reverse-biased and does not conduct. Thus,the output pulses $A$ and $C$ are produced by diode $1$.
During the negative half-cycle of the input $AC$ signal,diode $2$ becomes forward-biased and conducts,while diode $1$ is reverse-biased. Thus,the output pulses $B$ and $D$ are produced by diode $2$.
Therefore,the contribution to the output voltage from diode $2$ corresponds to $B$ and $D$.

(C) The given curve represents a full-wave rectified output current. The current $I$ varies as $I = I_0 \sin(\omega t)$ for the duration of each half-cycle.
To find the average value $(I_{avg})$ over one complete cycle of the rectified output (which has a period of $T/2 = \pi/\omega$),we integrate the current over one half-cycle and divide by the time interval:
$I_{avg} = \frac{1}{T/2} \int_0^{T/2} I_0 \sin(\omega t) dt$
$I_{avg} = \frac{\omega}{\pi} \int_0^{\pi/\omega} I_0 \sin(\omega t) dt$
$I_{avg} = \frac{\omega I_0}{\pi} \left[ -\frac{\cos(\omega t)}{\omega} \right]_0^{\pi/\omega}$
$I_{avg} = \frac{I_0}{\pi} [-\cos(\pi) + \cos(0)]$
$I_{avg} = \frac{I_0}{\pi} [1 + 1] = \frac{2I_0}{\pi}$

(B) For proper demodulation in an $AM$ detector,the time constant $RC$ must satisfy the condition: $\frac{1}{f_c} \ll RC < \frac{1}{f_m}$,where $f_c$ is the carrier frequency and $f_m$ is the message frequency.
First,calculate the period of the carrier signal: $\frac{1}{f_c} = \frac{1}{100 \times 10^3 \ Hz} = 10^{-5} \ s$.
Next,calculate the time constant of the circuit: $RC = (10^3 \ \Omega) \times (10 \times 10^{-12} \ F) = 10^{-8} \ s$.
Comparing the two,we see that $RC = 10^{-8} \ s$ is much smaller than $\frac{1}{f_c} = 10^{-5} \ s$.
Since the condition $\frac{1}{f_c} \ll RC$ is not satisfied,the circuit cannot effectively detect the $100 \ kHz$ carrier signal.

(A) rectifier converts $AC$ voltage into a pulsating $DC$ voltage. This pulsating $DC$ contains both $DC$ and $AC$ components (ripples). $A$ filter circuit,typically consisting of capacitors and inductors,is used to block or bypass the $AC$ ripples,thereby allowing only the $DC$ component to pass through to the load. Thus,the filter circuit removes $AC$ components from the rectified output.

(D) full-wave rectifier converts both halves of the input $AC$ cycle into a unidirectional output.
However,the output is not a steady $DC$ voltage like that from a battery.
It consists of a series of pulses that are always in the same direction,which is why it is referred to as a unidirectional pulsating voltage.
To obtain pure $DC$,a filter circuit is typically used after the rectifier.

(A) $P-N$ junction diode allows current to flow in only one direction (forward bias) and blocks it in the opposite direction (reverse bias). This property of unidirectional conduction makes it suitable for converting alternating current $(AC)$ into direct current $(DC)$. This device is known as a rectifier.

(D) half-wave rectifier uses a single diode in series with the load to convert $AC$ to $DC$.
In the provided options,the circuit in image $166-d103$ shows a transformer connected to a single diode in series with the load,which is the standard configuration for a half-wave rectifier.
Other options show either full-wave configurations or incorrect diode placements.

(B) In a full-wave rectifier, the circuit utilizes both halves of the input $AC$ cycle.
For each cycle of the input $AC$ signal, the output produces two pulses (one for the positive half-cycle and one for the negative half-cycle).
Therefore, the output frequency $(f_{out})$ is twice the input frequency $(f_{\in})$.
Given $f_{\in} = 50 \ Hz$.
$f_{out} = 2 \times f_{\in} = 2 \times 50 \ Hz = 100 \ Hz$.

(C) The given circuit is a bridge rectifier configuration.
When an $AC$ input is applied between terminals $A$ and $C$:
$1$. During the positive half-cycle,if $A$ is positive and $C$ is negative,the diodes connected to $A$ and $C$ will conduct such that current flows through the load connected between $B$ and $D$.
$2$. During the negative half-cycle,if $C$ is positive and $A$ is negative,the other pair of diodes will conduct,again ensuring that current flows through the load between $B$ and $D$ in the same direction.
Thus,the circuit acts as a full-wave rectifier,converting the $AC$ input into a pulsating $DC$ output across terminals $B$ and $D$.

(C) In a full-wave rectifier, the output is obtained for both the positive and negative half-cycles of the input alternating current $(AC)$.
Since the output pulses twice for every single cycle of the input $AC$, the frequency of the ripple in the output is double the input frequency.
Given input frequency $f_{in} = 50 \, Hz$.
Therefore, the fundamental ripple frequency $f_{out} = 2 \times f_{in} = 2 \times 50 \, Hz = 100 \, Hz$.

(D) During the positive half-cycle of the input voltage $(0 \leq \omega t \leq \pi)$,the diode is forward-biased. Since an ideal diode has zero resistance in forward bias,the voltage across it is zero.
During the negative half-cycle of the input voltage $(\pi \leq \omega t \leq 2\pi)$,the diode is reverse-biased and no current flows through it. Therefore,the input voltage appears across the diode.
Thus,the waveform shown in option $D$ will be obtained across the diode.

(D) The circuit shown is a full-wave rectifier. The anodes of both diodes $D_1$ and $D_2$ are connected to the transformer terminals $A$ and $B$ respectively,while the cathodes are connected together to one end of the load resistor $R_L$. The center tap $CT$ is connected to the other end of $R_L$.
When terminal $A$ is positive with respect to $CT$,diode $D_1$ is forward-biased,and current flows through $R_L$.
When terminal $B$ is positive with respect to $CT$,diode $D_2$ is forward-biased,and current flows through $R_L$ in the same direction.
However,in this specific circuit diagram,the diodes are oriented such that they conduct when the terminals $A$ and $B$ are negative with respect to $CT$. Thus,the current flows through $R_L$ in a direction that produces a negative voltage across it. Therefore,the output waveform shows full-wave rectification in the negative direction,which corresponds to option $D$.

(C) For a half-wave rectifier,the peak voltage $V_m$ is given by $V_m = \sqrt{2} \times V_{rms}$.
Given $V_{rms} = 220 \ V$,so $V_m = 1.414 \times 220 \approx 311.13 \ V$.
The average $DC$ voltage $V_{dc}$ is given by $V_{dc} = \frac{V_m}{\pi}$.
$V_{dc} = \frac{311.13}{3.14} \approx 99.08 \ V \approx 99 \ V$.
The average $DC$ current $I_{dc}$ is given by $I_{dc} = \frac{V_{dc}}{R_L}$.
Given $R_L = 1 \ k\Omega = 1000 \ \Omega$.
$I_{dc} = \frac{99 \ V}{1000 \ \Omega} = 0.099 \ A = 99 \ mA$.
Thus,the average $DC$ voltage is $99 \ V$ and the average $DC$ current is $99 \ mA$.

(C) The efficiency of a half-wave rectifier is given by the formula: $\eta = 40.6 \left( \frac{R_L}{R_d + R_L} \right) \%$.
Given: $R_L = 2 \, k\Omega$ and $R_d = 2 \, k\Omega$.
Substituting the values into the formula:
$\eta = 40.6 \left( \frac{2 \, k\Omega}{2 \, k\Omega + 2 \, k\Omega} \right) \%$.
$\eta = 40.6 \left( \frac{2}{4} \right) \%$.
$\eta = 40.6 \times 0.5 = 20.3 \%$.
Thus, the rectification efficiency is $20.3 \%$.

(A) The circuit consists of a $P-N$ junction diode in series with a load resistor $R_L$.
When the input signal is positive $(+5 \ V)$,the diode is forward-biased and acts as a closed switch (assuming an ideal diode). Thus,the entire input voltage appears across the load resistor $R_L$.
When the input signal is negative $(-5 \ V)$,the diode is reverse-biased and acts as an open switch. Thus,no current flows through $R_L$,and the output voltage across $R_L$ is $0 \ V$.
Therefore,the output signal is a square wave that varies between $0 \ V$ and $+5 \ V$.

(B) For a half-wave rectifier,the average or $DC$ output voltage $(V_{DC})$ is given by the formula:
$V_{DC} = \frac{V_m}{\pi}$
where $V_m$ is the peak (maximum) output voltage.
Given that $V_m = 10 \ V$,
$V_{DC} = \frac{10}{\pi} \ V$.

(A) full-wave rectifier circuit uses a center-tapped transformer and two diodes. The two diodes are connected to the two ends of the secondary coil of the transformer. The load resistor $R_L$ is connected between the center tap of the secondary coil and the common junction of the two diodes. During the first half cycle of the $AC$ input,one diode is forward-biased and conducts,while the other is reverse-biased. During the second half cycle,the roles are reversed. This ensures that current flows through the load resistor $R_L$ in the same direction during both half cycles. Option $A$ correctly depicts this configuration.

(C) The circuit shown is a half-wave rectifier circuit.
During the positive half-cycle of the input $AC$ voltage,the diode is forward-biased and conducts current through the resistor $R$.
During the negative half-cycle of the input $AC$ voltage,the diode is reverse-biased and does not conduct,so the current through the resistor $R$ is zero.
Therefore,the output current waveform across the resistor $R$ consists of only the positive half-cycles,which corresponds to the waveform shown in option $C$.

$(D)$ The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{30}{10 + 1500} = \frac{30}{1510} \approx 0.01987 \ A = 19.87 \ mA$.
The $D.C.$ load current is $I_{dc} = \frac{2I_m}{\pi} = \frac{2 \times 19.87 \ mA}{3.1416} \approx 12.65 \ mA$.
The $D.C.$ power output is $P_{dc} = I_{dc}^2 \times R_L = (12.65 \times 10^{-3})^2 \times 1500 \approx 240.1 \ mW$.
The $R.M.S.$ current for a full-wave rectifier is $I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{19.87}{\sqrt{2}} \approx 14.05 \ mA$.
The $A.C.$ power input is $P_{in} = I_{rms}^2 \times (R_f + R_L) = (14.05 \times 10^{-3})^2 \times (10 + 1500) = 1.974 \times 10^{-4} \times 1510 \approx 298.1 \ mW$.
Comparing with the given options, the closest values are $240.4 \ mW$ and $295.9 \ mW$.

(B) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L}$.
Given $V_m = 30 \ V$,$R_f = 10 \ \Omega$,and $R_L = 1500 \ \Omega$:
$I_m = \frac{30}{10 + 1500} = \frac{30}{1510} \approx 0.019868 \ A \approx 19.9 \ mA$.
For a full-wave rectifier,the average $(DC)$ current $I_{dc}$ is given by $I_{dc} = \frac{2 I_m}{\pi}$.
$I_{dc} = \frac{2 \times 19.9 \ mA}{3.14159} \approx 12.66 \ mA$.
The $RMS$ current $I_{rms}$ for a full-wave rectifier is given by $I_{rms} = \frac{I_m}{\sqrt{2}}$.
$I_{rms} = \frac{19.9 \ mA}{1.414} \approx 14.08 \ mA$.
Thus,the values are $19.9 \ mA, 12.66 \ mA, 14.08 \ mA$.

(A) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{25}{10 + 1000} = \frac{25}{1010} \approx 24.75 \ mA$.
The average current $I_{dc}$ for a half-wave rectifier is $I_{dc} = \frac{I_m}{\pi} = \frac{24.75}{3.14} \approx 7.88 \ mA$.
The $rms$ current $I_{rms}$ for a half-wave rectifier is $I_{rms} = \frac{I_m}{2} = \frac{24.75}{2} = 12.375 \ mA \approx 12.38 \ mA$.

(C) For a full-wave rectifier,the average $d.c.$ voltage is given by $V_{dc} = \frac{2V_m}{\pi} \approx 0.636 V_m$,where $V_m$ is the peak voltage of each secondary half-cycle.
Given the $rms$ voltage $V = 220 \ V$,the peak voltage is $V_m = V\sqrt{2} = 220\sqrt{2} \approx 311.13 \ V$.
Substituting this into the formula: $V_{dc} = 0.636 \times 311.13 \approx 198 \ V$.
Alternatively,$V_{dc} = \frac{2 \times (V\sqrt{2})}{\pi} = \frac{2\sqrt{2}}{\pi} V \approx 0.9 \times 220 = 198 \ V$.
The average $d.c.$ current is $I_{dc} = \frac{V_{dc}}{R_L} = \frac{198 \ V}{1000 \ \Omega} = 0.198 \ A = 198 \ mA$.

(D) For a full-wave rectifier,the peak voltage $V_m$ is given by $V_m = V_{rms} \times \sqrt{2}$.
Given $V_{rms} = 220 \ V$,so $V_m = 220 \times 1.414 = 311.13 \ V$.
The average $d.c.$ voltage is $V_{dc} = \frac{2V_m}{\pi} = 0.6366 \times 311.13 \approx 198.07 \ V$.
The ripple factor $r$ for a full-wave rectifier is $0.482$.
The ripple voltage $(rms)$ is given by $V_{r(rms)} = r \times V_{dc}$.
$V_{r(rms)} = 0.482 \times 198.07 \approx 95.47 \ V$.
Rounding to the nearest option,the value is $95.43 \ V$.

(C) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{25}{10 + 1000} = \frac{25}{1010} \approx 0.02475 \ A = 24.75 \ mA$.
The $DC$ current $I_{dc}$ is $I_{dc} = \frac{I_m}{\pi} = \frac{24.75}{\pi} \approx 7.88 \ mA$.
The $RMS$ current $I_{rms}$ for a half-wave rectifier is $I_{rms} = \frac{I_m}{2} = \frac{24.75}{2} = 12.375 \ mA$.
The ripple factor $\gamma$ is defined as $\gamma = \sqrt{\left( \frac{I_{rms}}{I_{dc}} \right)^2 - 1}$.
Substituting the values: $\gamma = \sqrt{\left( \frac{12.375}{7.88} \right)^2 - 1} = \sqrt{(1.57)^2 - 1} = \sqrt{2.46 - 1} = \sqrt{1.46} \approx 1.21$.

(D) For a half-wave rectifier,the peak voltage is $V_m = \sqrt{2} \times V_{rms} = \sqrt{2} \times 220 \approx 311.13 \ V$.
The $DC$ output voltage is $V_{dc} = \frac{V_m}{\pi} = \frac{311.13}{3.14} \approx 99 \ V$.
The ripple factor $r$ for a half-wave rectifier is $1.21$.
The ripple voltage $(rms)$ is given by $(V_r)_{rms} = r \times V_{dc}$.
Substituting the values: $(V_r)_{rms} = 1.21 \times 99 = 119.79 \ V$.

(B) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{25}{10 + 1000} = 24.75 \ mA$.
The $d.c.$ current is $I_{dc} = \frac{I_m}{\pi} = \frac{24.75}{3.14} \approx 7.88 \ mA$.
The $r.m.s.$ current is $I_{rms} = \frac{I_m}{2} = \frac{24.75}{2} = 12.38 \ mA$.
The $d.c.$ power output is $P_{dc} = I_{dc}^2 \times R_L = (7.88 \times 10^{-3})^2 \times 1000 \approx 62 \ mW$.
The $a.c.$ power input is $P_{ac} = I_{rms}^2 \times (R_f + R_L) = (12.38 \times 10^{-3})^2 \times (10 + 1000) \approx 155 \ mW$.

(D) The peak current $I_m$ is given by $I_m = \frac{V_m}{R_f + R_L} = \frac{25}{10 + 1000} = \frac{25}{1010} \approx 0.02475 \ A = 24.75 \ mA$.
The $DC$ current is $I_{dc} = \frac{I_m}{\pi} = \frac{24.75}{\pi} \approx 7.88 \ mA$.
The $RMS$ current is $I_{rms} = \frac{I_m}{2} = \frac{24.75}{2} = 12.375 \ mA$.
The $DC$ output power is $P_{dc} = I_{dc}^2 \times R_L = (7.88 \times 10^{-3})^2 \times 1000 \approx 0.0621 \ W = 62.1 \ mW$.
The $AC$ input power is $P_{ac} = I_{rms}^2 \times (R_f + R_L) = (12.375 \times 10^{-3})^2 \times (10 + 1000) \approx 0.1547 \ W = 154.7 \ mW$.
The efficiency $\eta$ is given by $\eta = \frac{P_{dc}}{P_{ac}} \times 100 = \frac{62.1}{154.7} \times 100 \approx 40.14 \ \%$.
Thus, the efficiency is approximately $40 \ \%$.

(C) For a half-wave rectifier,the average or $D.C.$ output voltage $(V_{dc})$ is given by the formula:
$V_{dc} = \frac{V_0}{\pi}$
where $V_0$ is the peak or maximum output voltage.
Given that $V_0 = 10 \ V$,we substitute this value into the formula:
$V_{dc} = \frac{10}{\pi} \ V$
Therefore,the $D.C.$ output voltage is $10/\pi \ V$.

(D) The given circuit is a half-wave rectifier circuit.
When the input signal is positive $(+5\,V)$,the diode is in forward bias and acts as a closed switch. Thus,the output voltage across $R_L$ is equal to the input voltage,which is $5\,V$.
When the input signal is negative $(-5\,V)$,the diode is in reverse bias and acts as an open switch. Thus,no current flows through $R_L$,and the output voltage across $R_L$ is $0\,V$.
Therefore,the output across $R_L$ will be a square wave with a peak voltage of $5\,V$ during the positive half-cycle and $0\,V$ during the negative half-cycle.

(C) The given waveform represents a full-wave rectified signal.
For a full-wave rectified signal,the average or $dc$ value is given by the formula:
$V_{dc} = \frac{2V_0}{\pi}$
From the graph,the peak voltage $V_0 = 6.28 \ V$.
Substituting the values:
$V_{dc} = \frac{2 \times 6.28}{3.14}$
$V_{dc} = 2 \times 2 = 4 \ V$.
Therefore,the correct option is $C$.