In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation.

In the circuit shown below, the switch is closed at t = 0. Applied vo

1.	In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation.
A.	-38.5±j1290
B.	38.5±j1290
C.	37.5±j1290
D.	-37.5±j1290
Answer» E.

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The complete solution of current obtained by substituting the values of c1 and c2 in the following equation is?i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).
The value of the c2 in the following equation is?i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).
The value of the c1 in the following equation is?i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).
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In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation.
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In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation.

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