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

${\mathcal {E}}\&{\mathcal {M}}$ (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 $V_{terminal}=\varepsilon -Ir_{eq}$ where $r_{eq}$ is the internal resistance and $\varepsilon$ 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: $\sum V=0$

▭ $V_{terminal}^{series}=\sum _{i=1}^{N}\varepsilon _{i}-I\sum _{i=1}^{N}r_{i}$ ▭ $V_{terminal}^{parallel}=\varepsilon -I\sum _{i=1}^{N}\left({\frac {1}{r_{i}}}\right)^{-1}$ where $r_{i}$ is internal resistance of each voltage source.
▭ Charging an RC (resistor-capacitor) circuit: $q(t)=Q\left(1-e^{-t/\tau }\right)$ and $I=I_{0}e^{-t/\tau }$ where $\tau =RC$ is RC time, $Q=\varepsilon C$ and $I_{0}=\varepsilon /R$.
▭ Discharging an RC circuit: $q(t)=Qe^{-t/\tau }$ and $I(t)=-{\tfrac {Q}{RC}}e^{-t/\tau }$

For quiz at QB/d_cp2.10

$V_{terminal}=\varepsilon -Ir_{eq}$ where $r_{eq}$=internal resistance and $\varepsilon$=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: $q(t)=Q\left(1-e^{t/\tau }\right)$ and $I=I_{0}e^{-t/\tau }$ where $\tau =RC$ is RC time, $Q=\varepsilon C$ and $I_{0}=\varepsilon /R$.

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