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Question

The reading of an ac ammeter connected to a half controlled rectifier with firing angle α = 0 will be :

This question was previously asked in
UGC NET 2015 Paper 3 History Question Paper (28-Jun-2015)
The correct answer is

\(\dfrac{I_{m}}{\sqrt{2}}\)

Two facts settle this: what an AC ammeter reads, and what a half-controlled bridge does at zero firing angle.

An AC ammeter is an RMS-reading instrument. Moving-iron and thermal instruments respond to the heating effect, which depends on the mean square of the current, so whatever the waveform they indicate

\(I_{rms}=\sqrt{\dfrac{1}{T}\int_{0}^{T}i^{2}(t)\,dt}\)

At α = 0 the half-controlled bridge behaves as an uncontrolled one. The thyristors are fired the instant they become forward biased, so they conduct exactly as diodes would, and each half-cycle of the supply is passed to the load in turn. The AC-side current is therefore a complete sinusoid of peak Im — no part of it is chopped away.

The RMS value of a full sine wave is the standard result :

\(I_{rms}=\sqrt{\dfrac{1}{2\pi}\int_{0}^{2\pi}\left(I_{m}\sin\theta\right)^{2}d\theta}=\dfrac{I_{m}}{\sqrt{2}}=0.707\,I_{m}\)

which is option 2.

OptionWhat it actually is
Im/2RMS of a half-wave rectified sine
Im/√2RMS of a full sine — correct
√2 ImDimensionally impossible — RMS can never exceed the peak
Im/2πNot a standard value; resembles a mean rather than an RMS

Option 3 can be discarded on sight: for any waveform the RMS value lies between zero and the peak, so a factor greater than one is impossible. Option 1 is the near miss for anyone who assumes only half the wave reaches the AC side — true for a half-wave circuit, but this is a bridge and both half-cycles are used.

What changes when α is increased. Each thyristor now conducts only from α to π, so the AC current becomes a chopped sinusoid and the RMS value falls:

\(I_{rms}(\alpha)=I_{m}\sqrt{\dfrac{1}{2\pi}\left(\pi-\alpha+\dfrac{\sin2\alpha}{2}\right)}\)

which reduces to \(I_{m}/\sqrt{2}\) at α = 0, confirming the answer.

Why the half-controlled bridge is used is that replacing two of the four thyristors with diodes halves the cost and the gate-drive complexity, and the resulting circuit cannot invert — it works in one quadrant only. Its freewheeling action also improves the input power factor compared with a fully controlled bridge at the same firing angle.

Hence, the ammeter reads Im/√2.

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Similar Questions

  1. In one quadrant converters such as half controlled bridge circuit or single phase circuit with fly wheel diodes for DC moter.

  2. In a two quadrant single phase SCR Drive armature current becomes continuous when :

  3. In a fully controlled three phase bridge without fly wheel operation the displacement factor and power factor is :


Important Questions from Phase Controlled Rectifiers

  1. If the input frequency of a bridge rectifier is 100 Hz, then the output frequency will be:

  2. Identify the expression:

    \(\rm \frac{V_{max}}{2\pi R_L}(1+\cos \alpha)\)

  3. Which mode is described when the anode is assigned a positive voltage, the gate is assigned a zero voltage disconnected and the cathode is assigned a negative voltage?
  4. The maximum firing angle that can be obtained by a pure resistive trigger circuit used in phase control circuit is:

  5. A 240 V single-phase AC supply is fed to a load resistor of 100 Ω through a thyristor. If the thyristor is fired at 90°, what is the power consumed by the load?

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