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I can obtain the transfer function in this RC circuit.

$$ \frac{V_{out}}{V_{in}} = \frac{1}{(RCs+1)} $$

When $V_{in}$ is set to $10V$, how can I get final value of $V_{out}$ at $t=∞$? I tried to use the final value theorem but I can't get the answer.

RC circuit

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    $\begingroup$ This is a low pass filter. This depends on the Vin frequency. Left of the cut off frequency Vout is base on the calculation. Right of cut off frequency Vout depends on the frequency, Hopefully this should help you. At larger frequencies Vout will be zero. $\endgroup$ – user8055 Sep 28 '20 at 15:17
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The final value theorem is for a signal, not a transfer function.

Use the transfer function to express the output signal $$ V_{\mathrm{out}}(s) = \frac{1}{RCs+1} V_{\mathrm{in}}(s),$$ with input $V_{\mathrm{in}}(s)$. Now, I assume that your input signal is a step-function $$ v_{\mathrm{in}}(t) = \begin{cases}0, \; \mathrm{for} \; t < 0 \\ 10, \; \mathrm{for} \; t \geq 0\end{cases},$$ with the corresponding Laplace transform $V_{\mathrm{in}}(s) = \mathcal{L} \left\lbrace v_{\mathrm{in}}(t) \right\rbrace = 10\frac{1}{s} $.

Now, we can apply the final value theorem $$ \begin{align} \lim_{s\to 0} s V_{\mathrm{out}}(s) &= \lim_{s\to 0} s \frac{1}{RCs+1} V_{\mathrm{in}}(s)\\ &= \lim_{s\to 0} s \frac{1}{RCs+1}\frac{10}{s} \\ &= \lim_{s\to 0} \frac{10}{RCs+1}\frac{s}{s} \\ &= \lim_{s\to 0} \frac{10}{RCs+1} \\ &= \frac{10}{RC\cdot 0+1}\\ &= 10. \end{align} $$

If your input signal is an impulse (Dirac delta function) $ v_{\mathrm{in}}(t) = \delta(t)$ the corresponding Laplace transform $V_{\mathrm{in}}(s) = \mathcal{L} \left\lbrace v_{\mathrm{in}}(t) \right\rbrace = 1 $ and the final value theorem is \begin{align} \lim_{s\to 0} s V_{\mathrm{out}}(s) &= \lim_{s\to 0} s \frac{1}{RCs+1} V_{\mathrm{in}}(s)\\ &= \lim_{s\to 0} s \frac{1}{RCs+1} 1 \\ &= \lim_{s\to 0} \frac{s}{RCs+1} \\ &= \lim_{s\to 0} \frac{0}{RC\cdot0+1} \\ &= 0 \end{align} Which we know is correct since the system is stable because the pole $p = -\frac{1}{RC} < 0$ for $R$, $C > 0$.

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