Question Solved1 Answer A series circuit has a capacitor of 0.25 x 10-6 F, a resistor of 5 x 103.12, and an inductor of 1 H. The initial charge on the capacitor is zero. If a 24-volt battery is connected to the circuit and the circuit is closed at t = 0, determine the charge on the capacitor at any time t. Q(t) = coulombs Determine the charge on the capacitor at t = 0.001 s, and at t = 0.01 s. (If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.) Q(0.001) coulombs Q(0.01) coulombs = Determine the limiting charge as t →0. Q(t) – coulombs as t →

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Transcribed Image Text: A series circuit has a capacitor of 0.25 x 10-6 F, a resistor of 5 x 103.12, and an inductor of 1 H. The initial charge on the capacitor is zero. If a 24-volt battery is connected to the circuit and the circuit is closed at t = 0, determine the charge on the capacitor at any time t. Q(t) = coulombs Determine the charge on the capacitor at t = 0.001 s, and at t = 0.01 s. (If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.) Q(0.001) coulombs Q(0.01) coulombs = Determine the limiting charge as t →0. Q(t) – coulombs as t →
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Transcribed Image Text: A series circuit has a capacitor of 0.25 x 10-6 F, a resistor of 5 x 103.12, and an inductor of 1 H. The initial charge on the capacitor is zero. If a 24-volt battery is connected to the circuit and the circuit is closed at t = 0, determine the charge on the capacitor at any time t. Q(t) = coulombs Determine the charge on the capacitor at t = 0.001 s, and at t = 0.01 s. (If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.) Q(0.001) coulombs Q(0.01) coulombs = Determine the limiting charge as t →0. Q(t) – coulombs as t →
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Voltage is inductor V_(L)=L(d^(2)q)/(dt^(2))Voltage in ressitor V_(R)=R(dq)/(dt)Vullage in Copacitor V_(C)=q//cBy kirchoffs 2^("hed ") lawL(d^(2)q)/(dt^(2))+R(dq)/(dt)+(q)/(c)=E(t)Given L=14quad R=5000chmquad C=0.25 xx10^(-6)fquad E(t)=R_(k){:[:.(d^(2)q)/(dt^(2))+5000(dq)/(dt)+(q)/(25 xx10^(-6))=12],[(d^(2)q)/(dt^(2))+5000(dq)/(dt)+4000000 q=12],[{:m^(2)+5000 m+4000000=0" (chavacteristic "q^(4))],[(m_(1)+4000)(m+1000)=0],[:.q_(c)=c_(1)e^(-4000 t)+c_(2)e^(-1000 t)]:}If Q(t)=A quadQ^(')(t)=0quadQ^('')(t)=0Substitutug in onginal eqh we ... See the full answer