Question
How many time constants will elapse before the energy stored in the capacitor reaches half of its equilibrium value in a charging RC circuit?
✓
Answer
$\text{q}=\text{Q}\big(1-\text{e}^{-\text{n}}\big)$
$\frac{1}{2}\frac{\text{Q}^2}{\text{C}}$ = Initial value; $\frac{1}{2}\frac{\text{q}^2}{\text{c}}$ = Final value
$\frac{1}{2}\frac{\text{q}^2}{\text{c}}\times2=\frac{1}{2}\frac{\text{Q}^2}{\text{C}}$
$\Rightarrow\text{q}^2=\frac{\text{Q}^2}{2}\Rightarrow\text{q}=\frac{\text{Q}}{\sqrt{2}}$
$\frac{\text{Q}}{\sqrt{2}}=\text{Q}(1-\text{e}^{\text{n}})$
$\Rightarrow\frac{1}{\sqrt{2}}=1-\text{e}^{-\text{n}}\Rightarrow\text{e}^{-\text{n}}=1-\frac{1}{\sqrt{2}}$
$\Rightarrow\text{n}=\log\Big(\frac{\sqrt{2}}{\sqrt{2}-1}\Big)=1.22$
$\frac{1}{2}\frac{\text{Q}^2}{\text{C}}$ = Initial value; $\frac{1}{2}\frac{\text{q}^2}{\text{c}}$ = Final value
$\frac{1}{2}\frac{\text{q}^2}{\text{c}}\times2=\frac{1}{2}\frac{\text{Q}^2}{\text{C}}$
$\Rightarrow\text{q}^2=\frac{\text{Q}^2}{2}\Rightarrow\text{q}=\frac{\text{Q}}{\sqrt{2}}$
$\frac{\text{Q}}{\sqrt{2}}=\text{Q}(1-\text{e}^{\text{n}})$
$\Rightarrow\frac{1}{\sqrt{2}}=1-\text{e}^{-\text{n}}\Rightarrow\text{e}^{-\text{n}}=1-\frac{1}{\sqrt{2}}$
$\Rightarrow\text{n}=\log\Big(\frac{\sqrt{2}}{\sqrt{2}-1}\Big)=1.22$
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