Question
A $100\mu\text{F}$ capacitor is joined to a 24V battery through a $1.0\text{M}\Omega$ resistor. Plot qualitative graphs (a) between current and time for the first 10 minutes and (b) between charge and time for the same period.
✓
Answer
Time constant $\text{RC} = 1 \times 10^6 \times 100 \times 10^6 = 100 \text{sec}$
I = Current $=\frac{\text{dq}}{\text{dt}}=\text{VC}.(-)\text{e}^{\frac{-\text{t}}{\text{RC}}},\frac{-1}{\text{RC}}$
$=\frac{\text{V}}{\text{R}}\text{e}^{\frac{-\text{t}}{\text{RC}}}=\frac{\text{V}}{\text{R}\cdot\text{e}^{\frac{\text{t}}{\text{RC}}}}=\frac{24}{10^6}\times\frac{1}{\text{e}^{\frac{\text{t}}{100}}}$
$=24\times10^{-6}\frac{1}{\text{e}^{\frac{\text{t}}{100}}}$
$\text{t}=10\text{min},600\text{sec}.$
$\text{Q}=24\times10+-4\times\big(1-\text{e}^{-6}\big)=23.99\times10^{-4}$
$\text{I}=\frac{24}{10^6}\times\frac{1}{\text{e}^6}=5.9\times10^{-8}\text{Amp}.$
- $\text{q}=\text{VC}\Big(1-\text{e}^{\frac{-\text{t}}{\text{CR}}}\Big)$
I = Current $=\frac{\text{dq}}{\text{dt}}=\text{VC}.(-)\text{e}^{\frac{-\text{t}}{\text{RC}}},\frac{-1}{\text{RC}}$
$=\frac{\text{V}}{\text{R}}\text{e}^{\frac{-\text{t}}{\text{RC}}}=\frac{\text{V}}{\text{R}\cdot\text{e}^{\frac{\text{t}}{\text{RC}}}}=\frac{24}{10^6}\times\frac{1}{\text{e}^{\frac{\text{t}}{100}}}$
$=24\times10^{-6}\frac{1}{\text{e}^{\frac{\text{t}}{100}}}$
$\text{t}=10\text{min},600\text{sec}.$
$\text{Q}=24\times10+-4\times\big(1-\text{e}^{-6}\big)=23.99\times10^{-4}$
$\text{I}=\frac{24}{10^6}\times\frac{1}{\text{e}^6}=5.9\times10^{-8}\text{Amp}.$
- $\text{q}=\text{VC}\Big(1-\text{e}^\frac{-\text{t}}{\text{CR}}\Big)$
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