Assuming a completely discharged capacitor in the beginning, during the charging process of a capacitor, after $t = \tau$, give the voltage level in relation to the maximum voltage level $U_\tau = x \cdot U_{final}$. State a value for x.
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Step 1: The voltage across a charging capacitor as a function of time is given by: $U(t) = U_{final}(1 - e^{-t/RC})$ where: $U(t)$ is the voltage at time t $U_{final}$ is the final voltage across the capacitor $R$ is the resistance in the circuit $C$ is the Show more…
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A capacitor that has been charged to $2.0 \times 10^{5} \mathrm{~V}$ is allowed to discharge through a resistor. What will be the voltage across the capacitor after five time constants have elapsed? We know, via [Eq. (34.10)], that after $n$ time constants, $q=$ $\mathrm{q}_{\infty}(0.368)^{\mathrm{n}} .$ Because $v$ is proportional to $q$ (that is, $\left.v=\mathrm{q} / \mathrm{C}\right)$, we may write $$ u n_{=5}=\left(2.0 \times 10^{5} \mathrm{~V}\right)(0.368)^{5}=1.4 \mathrm{kV} $$
A capacitor that has been charged to $2.0 \times 10^{5} \mathrm{~V}$ is allowed to discharge through a resistor. What will be the voltage across the capacitor after five time constants have elapsed? We know (p. 374) that after $n$ time constants, $q=q_{\infty}(0.368)^{n}$. Because $v$ is proportional to $q$ (that is, $v=q / C$ ), we may write $$ v_{n=5}=\left(2.0 \times 10^{5} \mathrm{~V}\right)(0.368)^{5}=1.4 \mathrm{kV} $$
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