A d.c. voltmeter has a sensitivity of 1000 O\/volt. When it measures half full scale in 100V range, the current through the voltmeter will be:<\/p>\n\n\n\n
a. 100mA
b. 0.5mA
c. 50mA
d. 1mA<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n The correct answer is b. 0.5mA<\/strong>.<\/p>\n\n\n\n The problem involves a DC voltmeter<\/strong> with a sensitivity of 1000 ohms per volt<\/strong>. Sensitivity refers to the internal resistance of the voltmeter for each volt of full-scale reading. In other words, for every volt that the voltmeter measures, it behaves like it has a resistance of 1000 ohms.<\/p>\n\n\n\n Since the voltmeter has a sensitivity of 1000 ohms per volt<\/strong>, for any voltage reading, the resistance of the voltmeter is calculated as:<\/p>\n\n\n\n [ The voltmeter is set to measure a 100V range<\/strong>, and it measures half of the full scale<\/strong>. Therefore, the voltage being measured is:<\/p>\n\n\n\n [ From the sensitivity specification of 1000 ohms per volt, the internal resistance of the voltmeter when measuring 50V would be:<\/p>\n\n\n\n [ Next, we apply Ohm\u2019s law to calculate the current through the voltmeter. Assuming the voltmeter is connected in parallel with the voltage source, the current through the voltmeter is given by:<\/p>\n\n\n\n [ So, the current through the voltmeter when it reads half of its full-scale value (50V on a 100V range) is 1mA<\/strong>. This matches option d. 1mA<\/strong>.<\/p>\n\n\n\n Despite the confusion in the wording of the question, the correct analysis leads to the answer that the current through the voltmeter is 1mA<\/strong> when it measures 50V on the 100V range. This corresponds to answer d. 1mA<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" A d.c. voltmeter has a sensitivity of 1000 O\/volt. When it measures half full scale in 100V range, the current through the voltmeter will be: a. 100mAb. 0.5mAc. 50mAd. 1mA The correct answer and explanation is : The correct answer is b. 0.5mA. Explanation: The problem involves a DC voltmeter with a sensitivity of 1000 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-209872","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209872","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=209872"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209872\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209872"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209872"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209872"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation:<\/h3>\n\n\n\n
Step 1: Determine the resistance of the voltmeter.<\/h4>\n\n\n\n
R_{vm} = 1000 \\, \\Omega\/\\text{V} \\times \\text{Voltage measured in volts}
]<\/p>\n\n\n\nStep 2: Understand the measurement situation.<\/h4>\n\n\n\n
V_{measured} = 100V \\times \\frac{1}{2} = 50V
]<\/p>\n\n\n\nStep 3: Calculate the internal resistance of the voltmeter.<\/h4>\n\n\n\n
R_{vm} = 1000 \\, \\Omega\/\\text{V} \\times 50V = 50,000 \\, \\Omega
]<\/p>\n\n\n\nStep 4: Calculate the current through the voltmeter.<\/h4>\n\n\n\n
I = \\frac{V}{R_{vm}} = \\frac{50V}{50,000 \\, \\Omega} = 0.001A = 1mA
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n