IDC = Average currentf0 = mains freqeucnyC = capacitorCalculation:Given f0 = 50 Hz, C = 680 μf, Vdc = 30 V, and RL = 220 Ω, we get:\({I_{dc}} = \frac{{{v_{dc}}}}{{{R_L}}} = \frac{{30}}{{220}}\)Idc = 0.136\({V_r} = \frac{{{I_{dc}}}}{{{f_{oc}}}} = \frac{{.136}}{{50 \times 680 \times {{10}^{ - 6}}}}\)\({V_r} = 4.01\;V\)

"> IDC = Average currentf0 = mains freqeucnyC = capacitorCalculation:Given f0 = 50 Hz, C = 680 μf, Vdc = 30 V, and RL = 220 Ω, we get:\({I_{dc}} = \frac{{{v_{dc}}}}{{{R_L}}} = \frac{{30}}{{220}}\)Idc = 0.136\({V_r} = \frac{{{I_{dc}}}}{{{f_{oc}}}} = \frac{{.136}}{{50 \times 680 \times {{10}^{ - 6}}}}\)\({V_r} = 4.01\;V\)

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Find the peak-to-peak ripple voltage of half-wave rectifier and filter circuit which has a 680 μF filter capacitor, an average output voltage of 30 V, and a 220Ω load resistance. The mains frequency is 50 Hz.

BITSAT Biotechnology and its Applications in BITSAT 10 months ago

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Concept:The Peak to peak ripple voltage of Half wave rectifies is given by:\({V_r} = \frac{{{I_{dc}}}}{{{f_{oc}}}}\)Where, IDC = Average currentf0 = mains freqeucnyC = capacitorCalculation:Given f0 = 50 Hz, C = 680 μf, Vdc = 30 V, and RL = 220 Ω, we get:\({I_{dc}} = \frac{{{v_{dc}}}}{{{R_L}}} = \frac{{30}}{{220}}\)Idc = 0.136\({V_r} = \frac{{{I_{dc}}}}{{{f_{oc}}}} = \frac{{.136}}{{50 \times 680 \times {{10}^{ - 6}}}}\)\({V_r} = 4.01\;V\)

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Posted on 18 Nov 2024, this text provides information on BITSAT related to Biotechnology and its Applications in BITSAT. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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