OpenStax University Physics/E&M/Direct-Current Circuits

$ℰ\mathrm{\&}ℳ$ (Vol. 2):    5:Electric Charges and Fields    6:Gauss's Law    7:Electric Potential    8:Capacitance    9:Current and Resistance    10:Direct-Current Circuits    11:Magnetic Forces and Fields    12:Sources of Magnetic Fields    13:Electromagnetic Induction    14:Inductance    15:Alternating-Current Circuits    16:Electromagnetic Waves

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Chapter 10

Direct-Current Circuits

Terminal voltage $V_{terminal}=\epsilon -I{r}_{eq}$ where $r_{eq}$ is the internal resistance and $\epsilon$ is the electromotive force.
▭ Resistors in series and parallel: $R_{series}=\sum _{i=1}^N{R}_i$ ▭ $R_{parallel}^{-1}=\sum _{i=1}^N{R}_i^{-1}$
▭ Kirchoff's rules. Loop:$\sum I_{in}=\sum I_{out}$ Junction: $\sum V=0$

▭ $V_{terminal}^{series}=\sum _{i=1}^N{\epsilon }_i-I\sum _{i=1}^N{r}_i$ ▭ $V_{terminal}^{parallel}=\epsilon -I\sum _{i=1}^N{\left(\frac{1}{r_i}\right)}^{-1}$ where $r_i$ is internal resistance of each voltage source.
▭ Charging an RC (resistor-capacitor) circuit: $q(t)=Q(1-e^{-t/\tau })$ and $I=I_0{e}^{-t/\tau }$ where $\tau =RC$ is RC time, $Q=\epsilon C$ and $I_0=\epsilon /R$.
▭ Discharging an RC circuit: $q(t)=Q{e}^{-t/\tau }$ and $I(t)=-\frac{Q}{RC}{e}^{-t/\tau }$

For quiz at QB/d_cp2.10

$V_{terminal}=\epsilon -I{r}_{eq}$ where $r_{eq}$=internal resistance and $\epsilon$=emf.

$R_{series}=\sum _{i=1}^N{R}_i$ and $R_{parallel}^{-1}=\sum _{i=1}^N{R}_i^{-1}$

Kirchhoff Junction:$\sum I_{in}=\sum I_{out}$ and Loop: $\sum V=0$

Charging an RC (resistor-capacitor) circuit: $q(t)=Q(1-e^{t/\tau })$ and $I=I_0{e}^{-t/\tau }$ where $\tau =RC$ is RC time, $Q=\epsilon C$ and $I_0=\epsilon /R$.

Discharging an RC circuit: $q(t)=Q{e}^{-t/\tau }$ and $I(t)=-\frac{Q}{RC}{e}^{-t/\tau }$