Electrical Engineering BMT & UFC › Circuit Analysis 1

Series Resistor

With your knowledge of meshes and nodes, you can already solve some problems in electrical engineering. If, for example, you have a motor as a load that needs a voltage of U_Motor = 3 V, but your voltage source outputs U₀ = 5 V, you have to reduce the voltage of the source for the load by 2 V. To do this, you use a series resistor. Assume the motor needs a current of I_Motor = 1 A.

The series resistor is connected between the source and the load. We call the voltage across the series resistor U_Vor. The mesh equation applies:

\[ \begin{gathered} U_0 = U_{\mathrm{Vor}} + U_{\mathrm{Motor}} \\[4pt] U_{\mathrm{Vor}} = U_0 - U_{\mathrm{Motor}} = 5\,\mathrm{V} - 3\,\mathrm{V} = 2\,\mathrm{V} \\[4pt] \text{Series connection: } I_{\mathrm{Motor}} = I_{\mathrm{Vor}} = I_0 = 1\,\mathrm{A} \end{gathered} \]

This means that the voltage and current at the series resistor are known and it can be dimensioned. You must use the voltage and current present at the component. So do not take the source voltage or the motor voltage, but the voltage across the series resistor. The following applies:

\[ R_{\mathrm{Vor}} = \frac{U_{\mathrm{Vor}}}{I_{\mathrm{Vor}}} = \frac{2\,\mathrm{V}}{1\,\mathrm{A}} = 2\,\Omega \]

Simulation