Calculate voltage, current resistance and power with Ohms law. Formulas for Resistors in series, Resistors in Parallel and Voltage divider

Number of Bands:

Ohms and Power law

Voltage(Volt):

$V = I \cdot R$

$V = \frac{P}{I}$

$V = \sqrt{P \cdot R}$

Current(Ampere):

$I = \frac{V}{R}$

$I = \frac{P}{V}$

$I = \sqrt{\frac{P}{R}}$

Resistance(Ohm):

$R = \frac{V}{I}$

$R = \frac{V^2}{P}$

$R = \frac{P}{I^2}$

Power(Watt):

$P = V \cdot I $

$P = I^2 \cdot R$

$P = \frac{V^2}{R} $

Resistors in series

The total resistance $R_{\text{total}}$ of resistors in series is the sum of their individual resistances:

$R_{\text{total}} = R_1 + R_2 + R_3 + \dots + R_n$

Resistors in Parallel

The total resistance $R_{\text{total}}$ of resistors in parallel is the reciprocal of the sum of their reciprocals:

$R_{\text{total}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}}$

Voltage divider

The output voltage $V_{\text{out}}$ of a voltage divider with two resistors $R_1$ and $R_2$ is:

$V_{\text{out}} = V_{\text{in}} \cdot \frac{R_2}{R_1 + R_2}$

Resistors math example

Values:

$ R_1 = 100\,\Omega $, $ R_2 = 220\,\Omega $ (in series)
$ R_3 = 470\,\Omega $ (in parallel with $ R_1 + R_2 $)
Supply Voltage $ U = 5\,\text{V} $

Total resistance ($ R_{\text{total}} $):

$ R_{\text{series}} = R_1 + R_2 $

$ = 100\,\Omega + 220\,\Omega $

$ = 320\,\Omega $

$ \frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_3} $

$ = \frac{1}{320} + \frac{1}{470} $

$ \approx 0.00525266 $

$ R_{\text{total}} = \frac{1}{0.00525266} $

$ \approx 190.38\,\Omega $

Total current ($ I_{\text{total}} $):

$ I_{\text{total}} = \frac{U}{R_{\text{total}}} $

$ = \frac{5\,\text{V}}{190.38\,\Omega} $

$ \approx 0.02626\,\text{A} $

$ = 26.26\,\text{mA} $

Voltage drops:

Across $ R_1 $ and $ R_2 $:
$ U_{\text{series}} = 5\,\text{V} $

$ U_1 = \frac{R_1}{R_{\text{series}}} \cdot U_{\text{series}} $

$ = \frac{100}{320} \cdot 5\,\text{V} $

$ \approx 1.56\,\text{V} $

$ U_2 = \frac{R_2}{R_{\text{series}}} \cdot U_{\text{series}} $

$ = \frac{220}{320} \cdot 5\,\text{V} $

$ \approx 3.44\,\text{V} $
Across $ R_3 $:
$ U_3 = 5\,\text{V} $

Current through each resistor:

Through $ R_1 $ and $ R_2 $:
$ I_{\text{series}} = \frac{U_{\text{series}}}{R_{\text{series}}} $

$ = \frac{5\,\text{V}}{320\,\Omega} $

$ \approx 15.63\,\text{mA} $
Through $ R_3 $:
$ I_3 = \frac{U_3}{R_3} $

$ = \frac{5\,\text{V}}{470\,\Omega} $

$ \approx 10.64\,\text{mA} $

Power dissipation:

Power in $ R_1 $:
$ P_1 = I_{\text{series}}^2 \cdot R_1 $

$ = (0.01563\,\text{A})^2 \cdot 100\,\Omega $

$ \approx 24.41\,\text{mW} $
Power in $ R_2 $:
$ P_2 = I_{\text{series}}^2 \cdot R_2 $

$ = (0.01563\,\text{A})^2 \cdot 220\,\Omega $

$ \approx 53.71\,\text{mW} $
Power in $ R_3 $:
$ P_3 = \frac{(U_3)^2}{R_3} $

$ = \frac{(5\,\text{V})^2}{470\,\Omega} $

$ \approx 53.19\,\text{mW} $

Total power dissipation ($ P_{\text{total}} $):

$ P_{\text{total}} = U \cdot I_{\text{total}} $

$ = 5\,\text{V} \cdot 0.02626\,\text{A} $

$ \approx 131.3\,\text{mW} $

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Resistors

Ohms and Power law

Resistors in series

Resistors in Parallel

Voltage divider