A diode whose internal resistance is 40 - is used to supply power to a 1k - load from a 110VRMS source supply- Then the dc load current will be

The diode will work as a half wave rectifier thus nbsp; I_dc = I_mπ R = 1000 + 40 = 1040 Ω Now, nbsp; nbsp; I_m = V_mR = 110×√(2)1040 = 149.53 mA ∴ ...

406

You visited us 406 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Diodes

A diode whose...

Question

$A d i o d e w h o s e i n t e r n a l r e s i s t a n c e i s$ $40 Ω$ $i s u s e d t o s u p p l y p o w e r t o a$ $1 k Ω$ $l o a d f r o m a 110 V (R M S) s o u r c e s u p p l y . T h e n t h e d c l o a d c u r r e n t w i l l b e$

47.61 mA

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

18.26 mA

No worries! We‘ve got your back. Try BYJU‘S free classes today!

42 mA

No worries! We‘ve got your back. Try BYJU‘S free classes today!

149.58 mA

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 47.61 mA
The diode will work as a half wave rectifier

thus $I_{d c} = \frac{I_{m}}{π}$

$R = 1000 + 40 = 1040 Ω$

Now,

$I_{m} = \frac{V_{m}}{R} = \frac{110 \times \sqrt{2}}{1040} = 149.53 m A$

$∴ I_{d c} = \frac{149.58}{π} = 47.61 m A$

Suggest Corrections

Similar questions

Q. A diode whose internal resistance is 40

Ω

is used to supply power to a 1 K

Ω

load from a 110 V (RMS) source supply. Then the dc load current will be

Q. A diode whose internal resistance is

20 Ω

is to supply power to a 1000

Ω

load from a 110 V (rms) source of supply. The total input power to the circuit is _________W.

Q. An energy source will supply a constant current into the load if its internal resistance is

Q. An alternating current generator has an internal resistance

R_{g}

and an internal reactance

X_{g}

. It is used to supply power to a passive load consisting of a resistance

R_{g}

and a reactance

X_{L}

. For maximum power to be delivered from the generator to the load, the value of

X_{L}

is equal to

Q. In a zener-regulated power supply a zener diode with

V_{z} = 6

volt is used for regulation. The load current is to be

4 m A

and the unregulated input is

10

volt. What should be the value of series resistor

R_{s}

if the current through diode is five times the load current?

Join BYJU'S Learning Program

A diode whose internal resistance is 40Ω is used to supply power to a 1kΩ load from a 110V(RMS) source supply. Then the dc load current will be

The correct option is A 47.61 mAThe diode will work as a half wave rectifier thus Idc=Imπ R=1000+40=1040Ω Now, Im=VmR=110×√21040=149.53mA ∴Idc=149.58π=47.61mA

$A d i o d e w h o s e i n t e r n a l r e s i s t a n c e i s$ $40 Ω$ $i s u s e d t o s u p p l y p o w e r t o a$ $1 k Ω$ $l o a d f r o m a 110 V (R M S) s o u r c e s u p p l y . T h e n t h e d c l o a d c u r r e n t w i l l b e$

The correct option is A 47.61 mA
The diode will work as a half wave rectifier

thus $I_{d c} = \frac{I_{m}}{π}$

$R = 1000 + 40 = 1040 Ω$

Now,

$I_{m} = \frac{V_{m}}{R} = \frac{110 \times \sqrt{2}}{1040} = 149.53 m A$

$∴ I_{d c} = \frac{149.58}{π} = 47.61 m A$