Current is the same in every part of a series circuit.
IF more components (resistors) are added, the effective current is reduced, but will still be the same everywhere (in total).
i.e. $I_{\text{tot}} = I_{1} = I_{2} = I_{3} =\dots$
For a single component in series with the energy source, $V\approx \text{emf}$.
IF more components are added in series then the sum of their individual drops in voltage adds up to the sum of the voltage drop across all of them.
i.e. $V_{\text{tot}} = V_{1} + V_{2} + V_{3}+\dots$
Since $I_{\text{tot}} = I_{1} = I_{2} = I_{3} =\dots$ and $V_{\text{tot}} = V_{1} + V_{2} + V_{3}+\dots$, and $V=IR$,
$\therefore V_{\text{tot}}=IR_{\text{tot}}=IR_{1} + IR_{2} + IR_{3}+\dots$
And since $I$ is the same anywhere,, we can divide by $I$.
$R_{\text{tot}} = \frac{V_{\text{tot}}}{I_{\text{tot}}}= R_{1} + R_{2} + R_{3} +\dots$
Current is divided among components in different branches
i.e. $I_{\text{tot}} = I_{1} + I_{2} + I_{3} +\dots$
Voltage is the same in each branch.
i.e. $V_{tot} = V_{1} = V_{2} = V_{3} =\dots$
Since $I_{\text{tot}} = I_{1} + I_{2} + I_{3} +\dots$ and $V_{\text{tot}} = V_{1} = V_{2} = V_{3}=\dots$, and $V=IR$,
$\therefore I_{\text{tot}}=\frac{V_{\text{tot}}}{R_{\text{tot}}}=\frac{V_{1}}{R_{1}}+\frac{V_{2}}{R_{2}}+\frac{V_{3}}{R_{3}}+\dots$
And since $V$ is the same anywhere, we can divide by $V$.
$\frac{1}{R_{\text{tot}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} +\dots$
Measure of the difficulty with which charge moves through a medium ($\Omega$).
Resulting in Ohm's Law: $V=IR$.
Ohmic vs non-ohmic devices
If a resistor has a voltage drop of $4.50V$ and a current of $2.00A$ flowing through it, what's its resistance?
