A coil of resistance 10 \u03a9 and inductance 47.7 mH is connected to a 200 V, 50 Hz supply. Calculate (a) the current drawn from the supply, (b) the power factor of the circuit. 5 A capacitor of 50 \u00b5F capacitance is connected in series with a resistance of 50 \u03a9 to a 200 V, 50 Hz supply. Calculate (a) the current drawn from the supply, (b) the voltage across the resistance, and (c) the voltage across the capacitance.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n A coil with resistance (R) of 10 \u03a9 and inductance (L) of 47.7 mH is connected to a 200 V, 50 Hz supply.<\/strong><\/p>\n\n\n\n (a) Current drawn from the supply:<\/strong><\/p>\n\n\n\n (b) Power factor (( \\text{pf} )):<\/strong><\/p>\n\n\n\n A capacitor of 50 \u00b5F and a resistance of 50 \u03a9 are connected to a 200 V, 50 Hz supply.<\/strong><\/p>\n\n\n\n (a) Current drawn from the supply:<\/strong><\/p>\n\n\n\n (b) Voltage across the resistance:<\/strong><\/p>\n\n\n\n [ (c) Voltage across the capacitance:<\/strong><\/p>\n\n\n\n [ In both problems, we used the concept of impedance in AC circuits, which combines resistance and reactance (inductive or capacitive). In Problem 1, the inductive coil’s impedance is dominated by its inductive reactance, causing a lagging power factor. In Problem 2, the capacitive circuit’s impedance depends on the balance between resistance and capacitive reactance, where the voltage across components is calculated using Ohm’s Law and the reactance formula.<\/p>\n","protected":false},"excerpt":{"rendered":" A coil of resistance 10 \u03a9 and inductance 47.7 mH is connected to a 200 V, 50 Hz supply. Calculate (a) the current drawn from the supply, (b) the power factor of the circuit. 5 A capacitor of 50 \u00b5F capacitance is connected in series with a resistance of 50 \u03a9 to a 200 V, […]<\/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-185157","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185157","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=185157"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185157\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185157"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185157"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Solution to Problem 1:<\/h3>\n\n\n\n
\n
[
X_L = 2 \\pi f L
]
Substituting (f = 50 \\, \\text{Hz}) and (L = 47.7 \\, \\text{mH} = 0.0477 \\, \\text{H}):
[
X_L = 2 \\pi (50)(0.0477) = 14.96 \\, \\Omega
]<\/li>\n\n\n\n
[
Z = \\sqrt{R^2 + X_L^2}
]
Substituting (R = 10 \\, \\Omega) and (X_L = 14.96 \\, \\Omega):
[
Z = \\sqrt{10^2 + 14.96^2} = \\sqrt{100 + 223.81} = \\sqrt{323.81} \\approx 18.0 \\, \\Omega
]<\/li>\n\n\n\n
[
I = \\frac{V}{Z}
]
Substituting (V = 200 \\, \\text{V}) and (Z = 18.0 \\, \\Omega):
[
I = \\frac{200}{18.0} \\approx 11.11 \\, \\text{A}
]<\/li>\n<\/ol>\n\n\n\n\n
[
\\text{pf} = \\frac{R}{Z}
]
Substituting (R = 10 \\, \\Omega) and (Z = 18.0 \\, \\Omega):
[
\\text{pf} = \\frac{10}{18.0} \\approx 0.556
]<\/li>\n<\/ol>\n\n\n\n
\n\n\n\nSolution to Problem 2:<\/h3>\n\n\n\n
\n
[
X_C = \\frac{1}{2 \\pi f C}
]
Substituting (f = 50 \\, \\text{Hz}) and (C = 50 \\, \\mu\\text{F} = 50 \\times 10^{-6} \\, \\text{F}):
[
X_C = \\frac{1}{2 \\pi (50)(50 \\times 10^{-6})} \\approx 63.66 \\, \\Omega
]<\/li>\n\n\n\n
[
Z = \\sqrt{R^2 + X_C^2}
]
Substituting (R = 50 \\, \\Omega) and (X_C = 63.66 \\, \\Omega):
[
Z = \\sqrt{50^2 + 63.66^2} = \\sqrt{2500 + 4053.63} \\approx \\sqrt{6553.63} \\approx 80.91 \\, \\Omega
]<\/li>\n\n\n\n
[
I = \\frac{V}{Z}
]
Substituting (V = 200 \\, \\text{V}) and (Z = 80.91 \\, \\Omega):
[
I = \\frac{200}{80.91} \\approx 2.47 \\, \\text{A}
]<\/li>\n<\/ol>\n\n\n\n
V_R = I R
]
Substituting (I = 2.47 \\, \\text{A}) and (R = 50 \\, \\Omega):
[
V_R = 2.47 \\times 50 \\approx 123.5 \\, \\text{V}
]<\/p>\n\n\n\n
V_C = I X_C
]
Substituting (I = 2.47 \\, \\text{A}) and (X_C = 63.66 \\, \\Omega):
[
V_C = 2.47 \\times 63.66 \\approx 157.3 \\, \\text{V}
]<\/p>\n\n\n\n
\n\n\n\nExplanation:<\/h3>\n\n\n\n