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Solved Examples · Example 7

Q.If an AC voltage V=240sin⁡314tV = 240 \sin 314t is applied to an SCR HWR. If it has firing angle 30∘^\circ, determine VdcV_{dc} and IdcI_{dc} when load resistance of 25 Ω\Omega is connected.

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[!TLDR]

Here Vm=240V_m = 240 V (the peak of 240sin⁡314t240\sin 314t); the half-wave-rectifier formula gives Vdc=71.26V_{dc} = 71.26 V and Idc=2.85I_{dc} = 2.85 A.

For a sinusoid written as V=Vmsin⁡ωt=240sin⁡314tV = V_m \sin\omega t = 240\sin 314t, the peak voltage is read directly as Vm=240V_m = 240 V (and ω=314\omega = 314 rad/s, i.e. a 50 Hz supply). For an SCR half-wave rectifier the average load voltage over the full cycle 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=30∘\alpha = 30^\circ and R=25 ΩR = 25\ \Omega:

Vdc=2402π(1+cos⁡30∘)=2402π×(1+0.866)=71.26 VV_{dc} = \frac{240}{2\pi}\left(1 + \cos 30^\circ\right) = \frac{240}{2\pi}\times(1 + 0.866) = 71.26\ \text{V} …

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