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 ∴ ...

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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

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18.26 mA

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42 mA

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149.58 mA

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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,

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

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

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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,

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

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