Question 18.18: A bridge rectifier uses four identical diodes having forward...

Given Transformer secondary resistance = 5 Ω
Secondary voltage V_{rms} = 30 V, I_{d.c.} = 200 mA
Since only two diodes of the bridge rectifier circuit will conduct during positive of negative half cycle of the input signal, the diode forward resistance r_{f} = 2 × 5 Ω = 10 Ω
We know that,

V_{d.c.} = \frac{2V_{m}}{\pi}– I_{d.c.} (r_{f} + r_{s}) where V_{m} =\sqrt{2} V_{rms} =\sqrt{2} × 30 V

Therefore, V_{d.c.} =\frac{2 ×\sqrt{2} × 30}{\pi} – 200 × 10^{–3} (10 +5) = 24 V

Ripple factor = \frac{rms\,value\,of\,ripple\,at\,the\,output}{average\, value\,of\,output\,voltage}

Therefore, 0.48 = \frac{rms\, value\, of\, ripple\, at \,the \,output}{24}
Hence, rms value of ripple at the output = 0.48 × 24 = 11.52 V