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A series circuit has a 0.1 henry inductor, a 1 ohm resistor and a 10% farad capacitor. There is an applied voltage of (o=o.si 3t). Find the steady-state solution Qp = 1 x 10-sin(t) - 3 x 10-6cos(31) Qp = - 3 x 10 * sin(3t) + 1 x 10-6cos(31) Qp = -3x10-sin(3t) + 1x 10 cos(31) Qp = 1 x 10-3sin(31) – 3 x 10-scos(3t) Question 8 of 10 Moving to another question will save this response.

Answer :

In a series circuit with a 0.1 H inductor, a 1 Ω resistor, and a 10% F capacitor, the steady-state solution is given by Qp = 1 × 10^(-sint) − 3 × 10^(-6cost).

A series circuit is a circuit configuration where the components are connected end-to-end, forming a single path for current flow. In this circuit, there is an inductor with an inductance of 0.1 H, a resistor with a resistance of 1 Ω, and a capacitor with a capacitance of 10% F (10% of 1 F).

The steady-state solution refers to the behavior of the circuit after it has reached a stable condition, where voltages and currents no longer change with time. The given equation Qp = 1 × 10^(-sint) − 3 × 10^(-6cost) represents the steady-state charge on the capacitor, Qp, as a function of time, t.

The equation consists of a sine and cosine term with different coefficients and arguments. The coefficients determine the amplitude of the sinusoidal waveform, and the arguments involve the time variable, t. By plugging in different values of t, the equation can be used to calculate the corresponding values of Qp.

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