Assuming the ideal diode, draw the output waveform for the circuit given in figure, explain the waveform.

0 6 Views | Posted 4 months ago
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  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    This is a Long Answer Type Questions as classified in NCERT Exemplar

    Explanation- When the input voltage is equal to or less than 5 V, diode will be revers biased. It will offer high resistance in comparison to resistance ( ) R in series. Now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above 5 V.

    If input voltage is more than + 5 V, diode will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the

    voltagebeyond+5Vwillappearacross R.

    When

    ...more

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V
Vishal Baghel

 F=μ0×24πi1i2dl

105=107×2×5×5d=10100

d=5×102m

d = 5 cm

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Vishal Baghel

A portion of the output power is returned back to the input.

P
Payal Gupta

g r 0 r R

g 1 r 2 r > R

A
alok kumar singh

Power gain = ( Δ i c Δ i b ) 2 × R 0 R i

= ( 1 0 × 1 0 3 1 0 0 × 1 0 6 ) 2 × 2 1

= 2 × 104 = x × 104

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A
alok kumar singh

Voltage gain = ΔlCΔlB×RCRB

5×103100×106×20.5

= 200

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