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Consider this circuit. Which is a second order RLC circuit.

RLC1

The problem asks for value of \$\phi\$ from the initial conditions.

RLC2

I took the first derivative of the equation by hand but as you'd see signs doesn't match with the solution of the problem.

RLC3

How did they find this answer from the equation above?

I suppose they neglected the C term for simplicity for current equation.

Update :

After solving the other part of the problem which asks for A I recognised that the differentiation is correct. But the solution would be \$\phi = \tan^{-1}(-\frac{\alpha}{\omega_d})\$

Only I couldn't be sure whether one of the sides of the triangle could be \$-\alpha\$

RLC4

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If: -

$$\omega_d \sin (\phi) = \alpha\cos(\phi)$$ Then: -

$$\dfrac{\sin (\phi)}{\cos(\phi)} = \dfrac{\alpha}{\omega_d}$$

Or: -

$$\tan(\phi) = \dfrac{\alpha}{\omega_d}$$

Can you take it from here?

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