A series RLC circuit has R = 5 W, L = 0.2 H and C = 50 \u00b5F. The applied voltage is 200 V. Find (i) resonant frequency (ii) Q-factor (iii) bandwidth (iv) upper and lower half-power frequencies (v) current at resonance (vi) current at half-power points (vii) voltage across inductance at resonance.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Given the following values for the RLC circuit:<\/p>\n\n\n\n The resonant frequency ( f_0 ) for an RLC circuit is given by the formula:<\/p>\n\n\n\n [ Substituting the values of ( L ) and ( C ):<\/p>\n\n\n\n [ [ So, the resonant frequency<\/strong> is approximately 50.5 Hz<\/strong>.<\/p>\n\n\n\n The quality factor (Q-factor) of the circuit is given by:<\/p>\n\n\n\n [ Where ( \\Delta f ) is the bandwidth. However, ( Q ) can also be expressed as:<\/p>\n\n\n\n [ Substituting the known values:<\/p>\n\n\n\n [ So, the Q-factor<\/strong> is approximately 12.65<\/strong>.<\/p>\n\n\n\n The bandwidth ( \\Delta f ) of the circuit is related to the Q-factor by:<\/p>\n\n\n\n [ Substituting the values:<\/p>\n\n\n\n [ So, the bandwidth<\/strong> is approximately 4.0 Hz<\/strong>.<\/p>\n\n\n\n The upper and lower half-power frequencies are given by:<\/p>\n\n\n\n [ Substituting the values:<\/p>\n\n\n\n [ So, the upper half-power frequency<\/strong> is approximately 52.5 Hz<\/strong> and the lower half-power frequency<\/strong> is approximately 48.5 Hz<\/strong>.<\/p>\n\n\n\n The current at resonance ( I_0 ) is given by:<\/p>\n\n\n\n [ Substituting the known values:<\/p>\n\n\n\n [ So, the current at resonance<\/strong> is 40 A<\/strong>.<\/p>\n\n\n\n At the half-power points, the voltage and current drop by ( \\frac{1}{\\sqrt{2}} ) of their peak values. The current at half-power frequencies can be calculated as:<\/p>\n\n\n\n [ So, the current at half-power points<\/strong> is approximately 28.3 A<\/strong>.<\/p>\n\n\n\n At resonance, the voltage across the inductor is equal to the current multiplied by the inductive reactance ( X_L ):<\/p>\n\n\n\n [ Substituting the known values:<\/p>\n\n\n\n [ The voltage across the inductance is:<\/p>\n\n\n\n [ So, the voltage across the inductance at resonance<\/strong> is 2540 V<\/strong>.<\/p>\n\n\n\n This analysis provides a comprehensive view of the circuit’s behavior, including resonance characteristics and the impact on current and voltage at various frequencies.<\/p>\n","protected":false},"excerpt":{"rendered":" A series RLC circuit has R = 5 W, L = 0.2 H and C = 50 \u00b5F. The applied voltage is 200 V. Find (i) resonant frequency (ii) Q-factor (iii) bandwidth (iv) upper and lower half-power frequencies (v) current at resonance (vi) current at half-power points (vii) voltage across inductance at resonance. The correct […]<\/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-198450","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198450","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=198450"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198450\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=198450"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=198450"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=198450"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
(i) Resonant Frequency<\/strong><\/h3>\n\n\n\n
f_0 = \\frac{1}{2\\pi \\sqrt{LC}}
]<\/p>\n\n\n\n
f_0 = \\frac{1}{2\\pi \\sqrt{0.2 \\times 50 \\times 10^{-6}}}
]<\/p>\n\n\n\n
f_0 \\approx \\frac{1}{2\\pi \\sqrt{0.00001}} = \\frac{1}{2\\pi \\times 0.00316} \\approx 50.5 \\, \\text{Hz}
]<\/p>\n\n\n\n(ii) Q-factor<\/strong><\/h3>\n\n\n\n
Q = \\frac{f_0}{\\Delta f}
]<\/p>\n\n\n\n
Q = \\frac{1}{R} \\sqrt{\\frac{L}{C}}
]<\/p>\n\n\n\n
Q = \\frac{1}{5} \\sqrt{\\frac{0.2}{50 \\times 10^{-6}}} \\approx \\frac{1}{5} \\sqrt{4000} \\approx \\frac{1}{5} \\times 63.25 = 12.65
]<\/p>\n\n\n\n(iii) Bandwidth<\/strong><\/h3>\n\n\n\n
\\Delta f = \\frac{f_0}{Q}
]<\/p>\n\n\n\n
\\Delta f = \\frac{50.5}{12.65} \\approx 4.0 \\, \\text{Hz}
]<\/p>\n\n\n\n(iv) Upper and Lower Half-Power Frequencies<\/strong><\/h3>\n\n\n\n
f_{\\text{upper}} = f_0 + \\frac{\\Delta f}{2}, \\quad f_{\\text{lower}} = f_0 – \\frac{\\Delta f}{2}
]<\/p>\n\n\n\n
f_{\\text{upper}} = 50.5 + \\frac{4.0}{2} = 52.5 \\, \\text{Hz}
]
[
f_{\\text{lower}} = 50.5 – \\frac{4.0}{2} = 48.5 \\, \\text{Hz}
]<\/p>\n\n\n\n(v) Current at Resonance<\/strong><\/h3>\n\n\n\n
I_0 = \\frac{V}{R}
]<\/p>\n\n\n\n
I_0 = \\frac{200}{5} = 40 \\, \\text{A}
]<\/p>\n\n\n\n(vi) Current at Half-Power Points<\/strong><\/h3>\n\n\n\n
I_{\\text{half}} = \\frac{I_0}{\\sqrt{2}} \\approx \\frac{40}{\\sqrt{2}} \\approx 28.3 \\, \\text{A}
]<\/p>\n\n\n\n(vii) Voltage Across Inductance at Resonance<\/strong><\/h3>\n\n\n\n
X_L = 2\\pi f_0 L
]<\/p>\n\n\n\n
X_L = 2\\pi \\times 50.5 \\times 0.2 = 63.5 \\, \\Omega
]<\/p>\n\n\n\n
V_L = I_0 \\times X_L = 40 \\times 63.5 = 2540 \\, \\text{V}
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n
\n