{"id":54361,"date":"2024-04-16T00:08:52","date_gmt":"2024-04-16T00:08:52","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=54361"},"modified":"2024-04-16T00:08:52","modified_gmt":"2024-04-16T00:08:52","slug":"the-e-m-f-induced-in-the-armature-of-a-shunt-generator-is-600-v-the-armature-resistance-is-0-1-ohm-if-the-armature-current-is-200-a-the-terminal-voltage-will-be","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-e-m-f-induced-in-the-armature-of-a-shunt-generator-is-600-v-the-armature-resistance-is-0-1-ohm-if-the-armature-current-is-200-a-the-terminal-voltage-will-be\/","title":{"rendered":"The e.m.f. induced in the armature of a shunt generator is 600 V. The armature resistance is 0.1 ohm. If the armature current is 200 A, the terminal voltage will be"},"content":{"rendered":"<p>[amp_mcq option1=&#8221;640 V&#8221; option2=&#8221;620 V&#8221; option3=&#8221;600 V&#8221; option4=&#8221;580 V&#8221; correct=&#8221;option3&#8243;]<!--more--><\/p>\n<p>The correct answer is $\\boxed{\\text{D}}$.<\/p>\n<p>The terminal voltage of a shunt generator is given by the following equation:<\/p>\n<p>$V_T = E_A &#8211; I_A R_A$<\/p>\n<p>where:<\/p>\n<ul>\n<li>$V_T$ is the terminal voltage<\/li>\n<li>$E_A$ is the induced emf<\/li>\n<li>$I_A$ is the armature current<\/li>\n<li>$R_A$ is the armature resistance<\/li>\n<\/ul>\n<p>In this case, we are given that $E_A = 600 \\text{ V}$, $I_A = 200 \\text{ A}$, and $R_A = 0.1 \\Omega$. Substituting these values into the equation, we get:<\/p>\n<p>$V_T = 600 \\text{ V} &#8211; 200 \\text{ A} \\cdot 0.1 \\Omega = 580 \\text{ V}$<\/p>\n<p>Therefore, the terminal voltage of the shunt generator is 580 V.<\/p>\n<p>Option A is incorrect because it is the induced emf, not the terminal voltage.<\/p>\n<p>Option B is incorrect because it is the induced emf plus the armature drop. The armature drop is the voltage drop across the armature resistance, and it is given by the following equation:<\/p>\n<p>$I_A R_A = 200 \\text{ A} \\cdot 0.1 \\Omega = 20 \\text{ V}$<\/p>\n<p>Therefore, the armature drop is 20 V, and the induced emf plus the armature drop is 600 V + 20 V = 620 V.<\/p>\n<p>Option C is incorrect because it is the induced emf.<\/p>\n<p>Option D is correct because it is the terminal voltage.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>[amp_mcq option1=&#8221;640 V&#8221; option2=&#8221;620 V&#8221; option3=&#8221;600 V&#8221; option4=&#8221;580 V&#8221; correct=&#8221;option3&#8243;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[742],"tags":[],"class_list":["post-54361","post","type-post","status-publish","format-standard","hentry","category-d-c-generators","no-featured-image-padding"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>The e.m.f. induced in the armature of a shunt generator is 600 V. 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The armature resistance is 0.1 ohm. If the armature current is 200 A, the terminal voltage will be","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/the-e-m-f-induced-in-the-armature-of-a-shunt-generator-is-600-v-the-armature-resistance-is-0-1-ohm-if-the-armature-current-is-200-a-the-terminal-voltage-will-be\/","og_locale":"en_US","og_type":"article","og_title":"The e.m.f. induced in the armature of a shunt generator is 600 V. The armature resistance is 0.1 ohm. 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