Question

A half-wave rectifier has a load resistance of 3.5 Kohm. If the diode and secondary of
the transformer have a total resistance of 800kohm. and the ac input voltage has 240 V
(peak value), determine: (i) peak, rms and average values of current through load
(1) DC power output (111) AC power input (iv) rectification efficiency
cu reforms Obtain expression​

Answer 1

Answer:

dc power output

Explanation:

mene ba choice kiya mujhe pata nhe

Answer 2

Answer:

The peak current is $4.242$×$10^{-4}$, the rms current is $3$ ×$10^{-4}$, the average value of current is $2.7$×$10^{-4}$A, DC power output is $145.8$×$10^{-4} W$, AC power input is $720$×$10^{-4} W$, and the rectification efficiency is 20.25%.

Explanation:

The values given in the question are;

Load resistance(Rl)=3.5×10³Ω

The total resistance of diode and transformer(Rt)=800×10³Ω

ac input voltage i.e. Vrms=240 V

The ac input voltage will be the rms voltage.

Now let's solve the question one by one.

(i) $Irms=Vrms/Resistance$   (1)

Resistance=Rl+Rt=(3.5+800)×10³Ω=803.5×10³Ω

$Irms=240/803.5$×$10^{3}$≈$3$×$10^{-4}$A

$Ipeak=\sqrt{2}$×$Irms$$=4.242$×$10^{-4}$A

$Iaverage=2Ipeak/$π$=2.7$×$10^{-4}$A

(ii)DC power output(Po)=Output direct current(Idc)×Output Voltage(Vdc)  (2)

$Idc=Ipeak/$π$=4.242$×$10^{-4}$$/$π$=1.35$×$10^{-4}$A

$Vdc=0.45Vrms=0.45$×$240=108V$

Now by putting the values of Idc and Vdc in equation (2) we get;

Po=$1.35$×$10^{-4}$×$108=145.8$×$10^{-4} W$

(iii)AC power input(Pi)=$Irms$×$Vrms$   (3)

Now by substituting the values of Irms and Vrms in equation (3) we get;

Pi=$3$×$10^{-4}$×$240=720$×$10^{-4} W$

(iv)

Rectification efficiency(RE)=

            dc power output$/$ac input power ×100

Or

Rectification efficiency(RE)=$Po/Pi$×$100$        (4)

By substituting the values of Po and Pi in equation (4) we will get the rectification efficiency;

RE=$145.8$×$10^{-4} /720$×$10^{-4}$ ×$100=0.2025$×$100=20.25%$%.

Hence, the peak current is $4.242$×$10^{-4} A$,  the rms current is 3×$10^{-4}$A, the average value of current is $2.7$×$10^{-4}$A, DC power output is $145.8$×$10^{-4} W$,

AC power input is $720$×$10^{-4} W$, and the rectification efficiency is 20.25%.