Question
How many time constants will elapse before the power delivered by a battery drops to half of its maximum value in an RC circuit?
✓
Answer
Power $=\text{CV}^2=\text{Q}\times\text{V}$
Now, $\frac{\text{QV}}{2}=\text{QV}\times\text{e}^{\frac{-\text{t}}{\text{RC}}}$
$\Rightarrow\frac{1}{2}=\text{e}^{\frac{-\text{t}}{\text{RC}}}$
$\Rightarrow\frac{\text{t}}{\text{RC}}=-\text{In}\ 0.5$
$\Rightarrow-(-0.69)=0.69.$
Now, $\frac{\text{QV}}{2}=\text{QV}\times\text{e}^{\frac{-\text{t}}{\text{RC}}}$
$\Rightarrow\frac{1}{2}=\text{e}^{\frac{-\text{t}}{\text{RC}}}$
$\Rightarrow\frac{\text{t}}{\text{RC}}=-\text{In}\ 0.5$
$\Rightarrow-(-0.69)=0.69.$
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