Dynamic Formulas for the Capacitor and for the Inductor

Charging the capacitor from the CV(=Vs) src. ::

The voltage drop on the series resistor $$V_R=V_S-V_C$$

$$t_{0_{↑}^{↑}→V_C}=-R·C·ln\left({1-\frac{V_C}{V_S}}\right)=R·C·ln\frac{V_S}{V_R}$$

$$t_{V_{1_C}{}_{↑}^{↑}→V_{2_C}}=R·C·ln\frac{V_S-V_{1_C}}{V_S-V_{2_C}}=R·C·ln\frac{V_{1_R}}{V_{2_R}}$$

Dis-charging the capacitor through a fixed value resistor ::

$$t_{V_{2_C}{}_{↓}^{↓}→V_{1_C}}=-R·C·ln\frac{V_{1_C}}{V_{2_C}}=R·C·ln\frac{V_{2_C}}{V_{1_C}}$$

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"Charging" the inductor from the CV(=Vs) src. ::

The peak current $$I_{MAX}=\frac{V_S}R$$

$$t_{0_{↑}^{↑}→I_L}=-\frac LR·ln\left({1-\frac{I_L}{I_{MAX}}}\right)$$

$$t_{I_{1_L}{}_{↑}^{↑}→I_{2_L}}=\frac LR·ln\frac{I_{MAX}-I_{1_L}}{I_{MAX}-I_{2_L}}$$

"Dis-charging" the inductor through a fixed value resistor ::

$$t_{I_{2_L}{}_{↓}^{↓}→I_{1_L}}=\frac LR·ln\frac{I_{2_L}}{I_{1_L}}$$

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about how to get linear I U ramps on the inductor and on the capacitor : simulation

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