Question #07b7f

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Question #07b7f

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Answer 1 · 2016-03-24T17:42:42

See steps below.

Explanation:

(a) Given expression is $V = V_{0} \cdot e^{- \frac{t}{R \cdot C}}$ $V=V_0cdote^(-t/(RcdotC))$
To make $t$ $t$ its subject.
Divide both sides with $V_{0}$ $V_0$ and take natural logarithm of both sides.
In the first step we obtain
$\frac{V}{V_{0}} = e^{- \frac{t}{R \cdot C}}$ $V/V_0=e^(-t/(RcdotC))$
In the second step
$ln (\frac{V}{V_{0}}) = {ln e}^{- \frac{t}{R \cdot C}}$ $ln (V/V_0)=ln e^(-t/(RcdotC))$ , eliminating $e$ $e$
$ln (\frac{V}{V_{0}}) = - \frac{t}{R \cdot C}$ $ln (V/V_0)=-t/(RcdotC)$ , now making $t$ $t$ its subject
$t = - (R \cdot C) \cdot ln (\frac{V}{V_{0}})$ $t=-(RcdotC)cdotln (V/V_0)$ , absorbing $- v e sign in ln$ $-ve " sign in " ln$ function
$t = R \cdot C \cdot ln (\frac{V_{0}}{V})$ $t=RcdotCcdotln (V_0/V)$ .

(b) For the second part follow same steps as above to obtain
$t = λ \cdot ln (\frac{N_{0}}{N})$ $t=lambdacdotln (N_0/N)$

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Question #07b7f

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