# V(t) = Vm cos(ωt) is applied to a half-wave rectifier. What is the RMS value of the output wave?

This question was previously asked in
SSC JE EE Previous Paper 9 (Held on: 29 Oct 2020 Evening)
View all SSC JE EE Papers >
1. $$\frac{{{I_m}}}{{2\pi }}$$
2. $$\frac{{{I_m}}}{{\pi }}$$
3. $$\frac{2{{I_m}}}{{\pi }}$$
4. $$\frac{{{I_m}}}{{2}}$$

Option 4 : $$\frac{{{I_m}}}{{2}}$$
Free
ST 1: Reasoning & Mental Ability
6801
15 Questions 15 Marks 15 Mins

## Detailed Solution

Concept:

The DC value or the average value of a waveform f(t) is given by the Area under the curve for one complete time period; i.e.,

$${f_{DC}} = \frac{1}{T}\mathop \smallint \nolimits_o^T f\left( t \right)dt$$

T = Time Period

The RMS or the effective value is given by:

$${f_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^T {f^2}\left( t \right)dt}$$

Calculation:

For the half-wave rectifier, the output is present is only for one half of the input signal and clipped for the other half.

Positive half-wave rectifier clips the negative half of the input signal and only a positive part of the input signal is present. Negative half-wave rectifier clips the positive half of the input signal and only negative part of the input signal is present In a half-wave rectifier circuit, the average value of output voltage is:

$${V_{avg}} = \frac{{{V_m}}}{\pi }$$

The RMS value of the output voltage is:

$${V_{rms}} = \frac{{{V_m}}}{2}$$

Where Vm is the maximum value of supply voltage Parameters Half wave Center Tap FWR Bridge FWR Type of Transformer Step Down Center tapped Step Down VDC $$\frac{V_m}{\pi}$$ $$\frac{2V_m}{\pi}$$ $$\frac{2V_m}{\pi}$$ IDC $$\frac{I_m}{\pi}$$ $$\frac{2I_m}{\pi}$$ $$\frac{2I_m}{\pi}$$ Vrms $$\frac{V_m}{2}$$ $$\frac{{{{\rm{V}}_{\rm{m}}}}}{{\sqrt 2 }}$$ $$\frac{{{{\rm{V}}_{\rm{m}}}}}{{\sqrt 2 }}$$ Ripple factor 1.21 0.48 0.48 PIV Vm 2Vm Vm O/P Frequency F 2f 2f Number of Diodes 1 2 4 Form factor 1.57 1.11 1.11 Crest factor 2 1.41 1.41
{"ft_lang_label":"__label__en","ft_lang_prob":0.63874245,"math_prob":0.99996805,"size":1564,"snap":"2021-31-2021-39","text_gpt3_token_len":563,"char_repetition_ratio":0.14102565,"word_repetition_ratio":0.060606062,"special_character_ratio":0.3887468,"punctuation_ratio":0.07070707,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99992526,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,4,null,4,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T08:13:19Z\",\"WARC-Record-ID\":\"<urn:uuid:a7cd389e-c370-4051-9734-deb91ae28d77>\",\"Content-Length\":\"122476\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98b5f021-1aa3-4cca-bbce-8458978a973a>\",\"WARC-Concurrent-To\":\"<urn:uuid:86c0afd4-6417-4f4e-bc24-8e78c1a47182>\",\"WARC-IP-Address\":\"172.67.30.170\",\"WARC-Target-URI\":\"https://testbook.com/question-answer/vt-vm-cost-is-applied-to-a-half-wave--5ffd7c8b3fd2d0dcbc3add32\",\"WARC-Payload-Digest\":\"sha1:FSBZ7JNMSGZQ6CEDDLZBMRCNDG4DYK2Y\",\"WARC-Block-Digest\":\"sha1:56BRU752NVN45KU6XKDPNWEY6QALH4J4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055601.25_warc_CC-MAIN-20210917055515-20210917085515-00681.warc.gz\"}"}