{"id":45290,"date":"2025-07-01T05:08:43","date_gmt":"2025-07-01T05:08:43","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=45290"},"modified":"2025-07-01T05:08:45","modified_gmt":"2025-07-01T05:08:45","slug":"use-voltage-division-to-find-the-steady-state-expression-for-vot-in-the-circuit-in","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/use-voltage-division-to-find-the-steady-state-expression-for-vot-in-the-circuit-in\/","title":{"rendered":"Use voltage division to find the steady-state expression for\u00a0vo(t)\u00a0in the circuit in"},"content":{"rendered":"\n
Use voltage division to find the steady-state expression for vo(t) in the circuit in<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
In this circuit, we are given the input voltage vg(t)=100cos\u2061(8000t)\u2009Vv_g(t) = 100 \\cos(8000t) \\, Vvg\u200b(t)=100cos(8000t)V, and we need to use voltage division to find the steady-state output voltage vo(t)v_o(t)vo\u200b(t) in the form: vo(t)=Vocos\u2061(\u03c9t+\u03d5)v_o(t) = V_o \\cos(\\omega t + \\phi)vo\u200b(t)=Vo\u200bcos(\u03c9t+\u03d5)<\/p>\n\n\n\n
where \u03c9=8000\u2009rad\/s\\omega = 8000 \\, \\text{rad\/s}\u03c9=8000rad\/s and the phase \u03d5\\phi\u03d5 lies between \u2212180\u2218-180^\\circ\u2212180\u2218 and 180\u2218180^\\circ180\u2218.<\/p>\n\n\n\n
Step 1: Impedance of the Components<\/h3>\n\n\n\n
The circuit has three components:<\/p>\n\n\n\n
    \n
  • Resistor (R):<\/strong> R=320\u2009\u03a9R = 320 \\, \\OmegaR=320\u03a9<\/li>\n\n\n\n
  • Inductor (L):<\/strong> L=200\u2009mH=0.2\u2009HL = 200 \\, \\text{mH} = 0.2 \\, HL=200mH=0.2H<\/li>\n\n\n\n
  • Capacitor (C):<\/strong> C=125\u2009nF=1.25\u00d710\u22127\u2009FC = 125 \\, \\text{nF} = 1.25 \\times 10^{-7} \\, \\text{F}C=125nF=1.25\u00d710\u22127F<\/li>\n<\/ul>\n\n\n\n
    We will first calculate the impedances of the inductor and the capacitor.<\/p>\n\n\n\n
    Inductive Impedance:<\/h4>\n\n\n\n
    The impedance of the inductor is given by: ZL=j\u03c9L=j(8000)(0.2)=j1600\u2009\u03a9Z_L = j \\omega L = j (8000)(0.2) = j 1600 \\, \\OmegaZL\u200b=j\u03c9L=j(8000)(0.2)=j1600\u03a9<\/p>\n\n\n\n
    Capacitive Impedance:<\/h4>\n\n\n\n
    The impedance of the capacitor is given by: ZC=1j\u03c9C=1j(8000)(1.25\u00d710\u22127)=1j0.001=\u2212j1000\u2009\u03a9Z_C = \\frac{1}{j \\omega C} = \\frac{1}{j (8000)(1.25 \\times 10^{-7})} = \\frac{1}{j 0.001} = -j 1000 \\, \\OmegaZC\u200b=j\u03c9C1\u200b=j(8000)(1.25\u00d710\u22127)1\u200b=j0.0011\u200b=\u2212j1000\u03a9<\/p>\n\n\n\n
    Step 2: Total Impedance of the Circuit<\/h3>\n\n\n\n
    Now, calculate the total impedance of the series circuit, which consists of the resistor RRR, inductor LLL, and capacitor CCC. The total impedance ZtotalZ_{\\text{total}}Ztotal\u200b is the sum of the individual impedances: Ztotal=R+ZL+ZC=320+j1600\u2212j1000=320+j600\u2009\u03a9Z_{\\text{total}} = R + Z_L + Z_C = 320 + j1600 – j1000 = 320 + j600 \\, \\OmegaZtotal\u200b=R+ZL\u200b+ZC\u200b=320+j1600\u2212j1000=320+j600\u03a9<\/p>\n\n\n\n
    Step 3: Voltage Division<\/h3>\n\n\n\n
    We use voltage division to find the output voltage. The voltage across the components in the series circuit is given by: vo(t)=vg(t)\u22c5ZoutZtotalv_o(t) = v_g(t) \\cdot \\frac{Z_{\\text{out}}}{Z_{\\text{total}}}vo\u200b(t)=vg\u200b(t)\u22c5Ztotal\u200bZout\u200b\u200b<\/p>\n\n\n\n
    Here, ZoutZ_{\\text{out}}Zout\u200b is the impedance seen by the output voltage, which is the combination of the impedance of the capacitor and resistor in series: Zout=ZC+R=\u2212j1000+320=320\u2212j1000\u2009\u03a9Z_{\\text{out}} = Z_C + R = -j1000 + 320 = 320 – j1000 \\, \\OmegaZout\u200b=ZC\u200b+R=\u2212j1000+320=320\u2212j1000\u03a9<\/p>\n\n\n\n
    Thus, the output voltage becomes: vo(t)=100cos\u2061(8000t)\u22c5320\u2212j1000320+j600v_o(t) = 100 \\cos(8000t) \\cdot \\frac{320 – j1000}{320 + j600}vo\u200b(t)=100cos(8000t)\u22c5320+j600320\u2212j1000\u200b<\/p>\n\n\n\n
    Step 4: Simplification and Expression for vo(t)v_o(t)vo\u200b(t)<\/h3>\n\n\n\n
    Now, simplify the above equation by calculating the magnitude and phase of the complex voltage divider:<\/p>\n\n\n\n
      \n
    1. Magnitude:<\/strong><\/li>\n<\/ol>\n\n\n\n
      The magnitude of the voltage ratio is: \u2223320\u2212j1000320+j600\u2223=3202+(\u22121000)23202+6002=102400+1000000102400+360000=1102400462400=1049.76679.53\u22481.544| \\frac{320 – j1000}{320 + j600} | = \\frac{\\sqrt{320^2 + (-1000)^2}}{\\sqrt{320^2 + 600^2}} = \\frac{\\sqrt{102400 + 1000000}}{\\sqrt{102400 + 360000}} = \\frac{\\sqrt{1102400}}{\\sqrt{462400}} = \\frac{1049.76}{679.53} \\approx 1.544\u2223320+j600320\u2212j1000\u200b\u2223=3202+6002\u200b3202+(\u22121000)2\u200b\u200b=102400+360000\u200b102400+1000000\u200b\u200b=462400\u200b1102400\u200b\u200b=679.531049.76\u200b\u22481.544<\/p>\n\n\n\n
        \n
      1. Phase:<\/strong><\/li>\n<\/ol>\n\n\n\n
        The phase is given by: \u03d5=arg\u2061(320\u2212j1000)\u2212arg\u2061(320+j600)\\phi = \\arg(320 – j1000) – \\arg(320 + j600)\u03d5=arg(320\u2212j1000)\u2212arg(320+j600) arg\u2061(320\u2212j1000)=tan\u2061\u22121(\u22121000320)=tan\u2061\u22121(\u22123.125)\u2248\u221272.3\u2218\\arg(320 – j1000) = \\tan^{-1}\\left(\\frac{-1000}{320}\\right) = \\tan^{-1}(-3.125) \\approx -72.3^\\circarg(320\u2212j1000)=tan\u22121(320\u22121000\u200b)=tan\u22121(\u22123.125)\u2248\u221272.3\u2218 arg\u2061(320+j600)=tan\u2061\u22121(600320)=tan\u2061\u22121(1.875)\u224862.1\u2218\\arg(320 + j600) = \\tan^{-1}\\left(\\frac{600}{320}\\right) = \\tan^{-1}(1.875) \\approx 62.1^\\circarg(320+j600)=tan\u22121(320600\u200b)=tan\u22121(1.875)\u224862.1\u2218<\/p>\n\n\n\n
        Thus, the total phase is: \u03d5=\u221272.3\u2218\u221262.1\u2218=\u2212134.4\u2218\\phi = -72.3^\\circ – 62.1^\\circ = -134.4^\\circ\u03d5=\u221272.3\u2218\u221262.1\u2218=\u2212134.4\u2218<\/p>\n\n\n\n
        Step 5: Final Expression for vo(t)v_o(t)vo\u200b(t)<\/h3>\n\n\n\n
        Using the magnitude and phase, the steady-state output voltage is: vo(t)=1.544\u22c5100cos\u2061(8000t\u2212134.4\u2218)v_o(t) = 1.544 \\cdot 100 \\cos(8000t – 134.4^\\circ)vo\u200b(t)=1.544\u22c5100cos(8000t\u2212134.4\u2218)<\/p>\n\n\n\n
        Thus, the steady-state expression for vo(t)v_o(t)vo\u200b(t) is: vo(t)=154.4cos\u2061(8000t\u2212134.4\u2218)\u2009Vv_o(t) = 154.4 \\cos(8000t – 134.4^\\circ) \\, \\text{V}vo\u200b(t)=154.4cos(8000t\u2212134.4\u2218)V<\/p>\n\n\n\n
        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
        Use voltage division to find the steady-state expression for vo(t) in the circuit in The Correct Answer and Explanation is: In this circuit, we are given the input voltage vg(t)=100cos\u2061(8000t)\u2009Vv_g(t) = 100 \\cos(8000t) \\, Vvg\u200b(t)=100cos(8000t)V, and we need to use voltage division to find the steady-state output voltage vo(t)v_o(t)vo\u200b(t) in the form: vo(t)=Vocos\u2061(\u03c9t+\u03d5)v_o(t) = V_o \\cos(\\omega t […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-45290","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45290","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=45290"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45290\/revisions"}],"predecessor-version":[{"id":45293,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/45290\/revisions\/45293"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=45290"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=45290"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=45290"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}