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Question Number 146667 by puissant last updated on 14/Jul/21

Answered by Olaf_Thorendsen last updated on 14/Jul/21

$$1.$$$${\mathrm{U}}_{\mathrm{C}}\left(t\right)={\mathrm{U}}_{\mathrm{R}}\left(t\right)={\mathrm{RI}}_{\mathrm{R}}\left(t\right)=\mathrm{R}(\mathrm{I}-{\mathrm{I}}_{\mathrm{C}}\left(t\right))$$$${\mathrm{Q}}_{\mathrm{C}}\left(t\right)={\mathrm{CU}}_{\mathrm{C}}\left(t\right)\mathrm{et}{\mathrm{I}}_{\mathrm{C}}\left(t\right)=\frac{d{\mathrm{Q}}_{\mathrm{C}}\left(t\right)}{dt}$$$$\frac{1}{\mathrm{C}}{\mathrm{Q}}_{\mathrm{C}}\left(t\right)=\mathrm{R}(\mathrm{I}-\frac{d{\mathrm{Q}}_{\mathrm{C}}\left(t\right)}{dt})$$$$\frac{d{\mathrm{Q}}_{\mathrm{C}}\left(t\right)}{dt}+\frac{1}{\mathrm{RC}}{\mathrm{Q}}_{\mathrm{C}}\left(t\right)=\mathrm{I}$$$$\mathrm{Posons}\tau =\mathrm{RC}$$$$\frac{d{\mathrm{Q}}_{\mathrm{C}}\left(t\right)}{dt}+\frac{1}{\tau }{\mathrm{Q}}_{\mathrm{C}}\left(t\right)=\mathrm{I}$$$${\mathrm{Q}}_{\mathrm{C}}\left(t\right)=\tau \mathrm{I}(1-e^{-\frac{t}{\tau }})$$$$$$$$2.$$$${\mathrm{U}}_{\mathrm{C}}\left(t\right)=\frac{1}{\mathrm{C}}{\mathrm{Q}}_{\mathrm{C}}\left(t\right)=\mathrm{RI}(1-e^{-\frac{t}{\tau }})$$$${\mathrm{I}}_{\mathrm{C}}\left(t\right)=\frac{d{\mathrm{Q}}_{\mathrm{C}}\left(t\right)}{dt}=\mathrm{I}{e}^{-\frac{t}{\tau }}$$

Commented by puissant last updated on 15/Jul/21

$$\mathrm{merci}\mathrm{monsieur}$$

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