The circuit shown in the following is used to control the brightness when turning on a small light bulb.
The circuit contains a voltage source $U=12 ~\rm{V}$, a switch $S_1$, a resistor of $R_1=20 ~\Omega$ and a capacitor of $C=100 ~\rm{µF}$.
The switch $S_2$ to an additional consumer $R_2$ will be considered to be open for the first tasks. At the moment $t_0=0 ~\rm{s}$ the switch $S_1$ is closed, the voltage across the capacitor is $u_c (t_0 )=0 ~\rm{V}$.
1. First do not consider the light bulb – it is not connected to the RC circuit.
Calculate the point of time $t_1$ when $u_c (t_1)=0.5\cdot U$.
So, here only R_1 and C gives the time constant: $\tau = R_1 \cdot C$
The following formula describes the time course of $u_C(t)$ which has to be $u_c (t_1)=0.5\cdot U$: \begin{align*} u_c (t) = U \cdot (1- e^{t/\tau}) = 0.5\cdot U \end{align*} It has to be rearranged to $t$ \begin{align*} (1- e^{t/\tau}) &= 0.5 \\ e^{t/\tau} &= 0.5 \\ t/\tau &= ln(0.5) \\ t &= \tau \cdot ln(0.5) \\ t &= R_1 \cdot C \cdot ln(0.5) \end{align*}
2. Calculate the overall energy dissipated by $R_1$ while charging the capacitor $0 ~\rm{V}$ to $12 ~\rm{V}$.
3. Now, consider the light bulb as a resistor of $R_\rm B=20 ~\Omega$, and ignore again the left side ($S_2$ is open).
The voltage across the capacitor is again $0 ~\rm{V}$ at the moment $t_0=0 ~\rm{s}$ when the switch $S_1$ is closed.
Calculate the voltage $u_c (t_2)$ across the capacitor at $t_2=1 ~\rm{ms}$ after closing the switch.
Hint: To solve this, first create an equivalent linear voltage source from $U$, $R_1$, and $R_\rm B$.
An equivalent linear voltage source can be given with $U$, $R_1$, and $R_\rm B$ as seen in yellow.
Therefore, the voltage of the equivalent linear voltage source is: $U_s = U \cdot {{R_\rm B}\over{R_1 + R_\rm B}} = 1/2 \cdot U$
The internal resistance is given by substituting the ideal voltage source with its resistance ($=0 ~\Omega$, short-circuit).
\begin{align*}
R_i &= R_1 || R_\rm B \\
&= 10 ~\Omega
\end{align*}
\begin{align*} u_c (t_2) &= U_s \cdot (1- e^{t_2/(R_i\cdot C)}) \\ &= {{1}\over{2}} \cdot U \cdot (1- e^{1~\rm{ms}/(10 ~\Omega \cdot 100 ~\rm{µF})}) \end{align*}
4. Explain (without calculation) how the situation in 3. would change once also $S_2$ is closed from the beginning on.