Home
Class 12
PHYSICS
In a half wave rectifier output is taken...

In a half wave rectifier output is taken across a `90 ohm` load resistor. If the resistance of diode in forward biased condition is `10 ohm`, the efficiency of rectification of `ac` power into `dc` power is.

A

`40.6%`

B

`81.2%`

C

`73.08%`

D

`36.54%`

Text Solution

Verified by Experts

The correct Answer is:
D
`eta=(0.406R_(L))/(r_(f)+R_(L))`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SEMICONDUCTOR ELECTRONICS

    NARAYNA|Exercise EXERCISE -1 (C.W) (TRANSISTORS)|8 Videos

  • SEMICONDUCTOR ELECTRONICS

    NARAYNA|Exercise EXERCISE -1 (C.W) (LOGIC GATES)|9 Videos

  • SEMICONDUCTOR ELECTRONICS

    NARAYNA|Exercise EXERCISE -1 (C.W) (JUNCTION DIODE)|5 Videos

  • SEMI CONDUCTOR DEVICES

    NARAYNA|Exercise Level-II (H.W)|36 Videos

  • WAVE OPTICS

    NARAYNA|Exercise Exercise - 4 Polarisation|9 Videos

Similar Questions

Explore conceptually related problems

A full wave p-n diode rectifier uses a load resistor of 1500Omega .No filter is used.The forward bias resistance of the diode is 10Omega ,the efficiency of the rectifier is

A full wave rectifier uses two diodes with a load resistance of 100 Omega . Each diode is having negligible forward resistance. Find the efficiency of this wave rectifier.

Knowledge Check

  • In a full wave rectifier output is taken across a load resistor of 800 ohm . If the resistance of diode in forward biased condition is 200 ohm , the efficiency of rectification of ac power into dc power is.

    A
    `64.96 %`
    B
    `40.6 %`
    C
    `81.2 %`
    D
    `80 %`

  • A 50V battery is connected across a 10 ohm resistor. The current is 4.5 amperes. The internal resistance of the battery is

    A
    Zero
    B
    `0.5` ohm
    C
    `1.1` ohm
    D
    `5.0` ohm

  • In a common emitter amplifier, using output resistance of 5000 ohm and input resistance of 2000 ohm, if the input signal voltage is 10mV and beta = 50 , calculate output volatge & power gain

    A
    `1.25V, 6250`
    B
    `3V, 6250`
    C
    `1.5V, 3050`
    D
    None of these

    • Similar Questions

    Explore conceptually related problems

    In half wave rectifier a p-n diode with internal resistance 20Omega is used. If the load resistance of 2kOmega is used in the circuit, then find the efficiency of this half wave rectifier.

    A p-n diode is used in a half wave rectifier with a load resistance of 1000 Omega . If the forward resistance (r_(f)) of diode is 10 Omega , calculate the efficiency of this half wave rectifier.

    A sinusoidal voltage of amplitude 25 volts and frequency 50 Hz is applied to a half wave rectifier using PN diode. No filter is used and the load resistor is 1000 Omega . The forward resistance R_(f) ideal diode is 10 Omega . Calculate. (i) Peak, average and rms values of load current (ii) d.c power output (ii) a.c power input (iv) % Rectifier efficiency (v) Ripple factor.

    A half wave rectifier provides DC output power of 30 W when 100 W AC power is supplied to it what will be the rectification efficiency and power efficiency

    In a common emitter amplifier, using output reisistance of 5000 ohm and input resistance fo 2000 ohm, if the peak value of input signal voltage is 10 m V and beta=50 . The power gain is

    Home
    Profile