#output voltage

Rectifiers, Filters and Applications

In a half-wave rectifier, the output voltage appears across the load during:

**Capacitor filter** purpose to reduce ripple, capacitor charges to peak during diode conduction and discharges through load during non-conduction, time constant τ = R C, larger RC less ripple, ripple reduced by using capacitor or LC filter, output of half-wave without filter is pulsating half sinusoids, full-wave without filter is two half sinusoids per cycle. A half-wave rectifier conducts only during the positive half-cycle of the AC input, allowing current through the load only in that period, producing a pulsating output for half the cycle. Substituting values gives Positive half-cycle, w

Ref: NCERT > Physics Book > Electronic Devices > Rectifiers, Filters and Applications

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Rectifiers, Filters and Applications

In a full-wave rectifier with a capacitor filter, the output voltage is closer to:

**Rectifier applications** diode must have reverse breakdown voltage higher than peak inverse voltage, centre-tap transformer provides two opposite phase voltages for full-wave, capacitor discharges through load R_L when diode off, drift current in junction is minority carrier motion due to field, diffusion due to gradient, diode conducts when forward biased anode positive. The capacitor charges to the peak voltage of the rectified output and discharges slowly, making the output voltage nearer to the peak value of the AC input. Substituting values gives Peak voltage, which matches expected beh

Ref: NCERT > Physics Book > Electronic Devices > Rectifiers, Filters and Applications

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Rotational EMF and AC Generator

In an AC generator, the frequency of the output voltage depends on which factor?

**AC generator** principle same as rotating coil, N=200 turns A=0.04 m² B=0.1 T f=50 Hz ω=2π×50=314 rad/s, e₀= N B A ω =200×0.1×0.04×314=251.2 V, emf e= e₀ sin ωt, frequency equals rotation frequency, maximum when plane parallel to field. The frequency equals the rotational speed of the coil in revolutions per second, as each full rotation produces one cycle of emf ( f = (ω/2π) ). Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l, M = e/(dI/dt) and U = ½ L I², result Rotational

Ref: NCERT > Physics Book > Electromagnetic Induction > Rotational EMF and AC Generator

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Magnetic Flux and Faraday's Laws of Induction

In an AC generator, increasing the number of turns in the coil affects what aspect of the output?

**Magnetic flux** Φ = B·A = B A cosθ, B magnetic field (T), A area (m²), θ angle between B and normal to area, unit Wb = T·m², Faraday's law induced emf e = -N dΦ/dt, N turns, negative sign Lenz's law indicating opposition, magnitude |e| = N |ΔΦ/Δt|, for 100 turns ΔΦ=0.03 Wb Δt=0.06 s e=100×0.03/0.06=50 V. The emf ( ε = N B A ω sinω t ) is directly proportional to the number of turns, so more turns increase the amplitude of the induced emf. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l

Ref: NCERT > Physics Book > Electromagnetic Induction > Magnetic Flux and Faraday's Laws of Induction

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