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A half-wave rectifier has an input voltage of 220 \mathrm{~V} \mathrm{rms}. If the stepdown transformer has a turns ratio of 4: 1, what is the peak load voltage? Ignore diode drop.

Option: 1

154.7 \mathrm{~V}


Option: 2

310 \mathrm{~V}


Option: 3

53 \mathrm{~V}


Option: 4

88 \mathrm{~V}


Answers (1)

best_answer

We need to convert the rms voltage to peak voltage using the relation

V_p=\sqrt{2} \times V_{\text {rms }}

\begin{aligned} V_p & =\sqrt{2} \times V_{\text {in }} \\ V_p & =\sqrt{2} \times 220, \mathrm{~V} \\ V_p & =220 \times 1.414 \\ V_p & =310.48, \mathrm{~V}_{\text {(approximately) }} \end{aligned}

Low voltage rectifiers require a step-down transformer to reduce the strength of \mathrm{AC} voltage.

                                                                  \frac{V_1}{V_2}=\frac{N_1}{N_2}

                                         \frac{N_1}{N_2}=\frac{4}{1}

\mathrm{V}_p at the output of stepdown transformer:


\mathrm{V}_p=\frac{210.48}{4} \approx 53 \mathrm{~V}

Posted by

vishal kumar

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