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Solved Examples · Example 5

Q.Determine VdcV_{dc} and IdcI_{dc} of SCR HWR. Given firing angle is 90∘^\circ and rms voltage of ac input to the rectifier is 230 V and load is 10 Ω\Omega.

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[!TLDR]

With Vm=2×230=325.2V_m = \sqrt{2}\times 230 = 325.2 V and cos⁡90∘=0\cos 90^\circ = 0, Vdc=51.76V_{dc} = 51.76 V and Idc=5.176I_{dc} = 5.176 A.

This SCR half-wave rectifier calculation is a standard Karnataka 2nd PUC Electronics exam numerical. In a half-wave rectifier controlled by RC triggering, the thyristor conducts from the firing angle α\alpha to π\pi once per cycle, so the average (dc) load voltage taken over the full 2π2\pi period is

Vdc=Vm2π(1+cos⁡α),Idc=VdcRV_{dc} = \frac{V_m}{2\pi}\left(1 + \cos\alpha\right), \qquad I_{dc} = \frac{V_{dc}}{R}

Given α=90∘\alpha = 90^\circ, Vrms=230V_{rms} = 230 V and R=10 ΩR = 10\ \Omega: …

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