MEXLE Nuggets

Voltage divider as voltage source

The voltage divider shown in figure 1 is in an unloaded state, as the entire current supplied by the power supply flows through the resistors $R_{\rm 1}$ and $R_{\rm 2}$ connected in series. A resistor parallel to $R_{\rm 2}$ loads the voltage divider. Set the voltage on the power supply to $12 ~\rm V$ and measure the exact voltage with a multimeter. Set up the measuring circuit shown in figure 1. For the connected load $R_{\rm L}$ = ${\rm 10} ~{\rm k\Omega}$ , the voltage divider represents a voltage source. Like any voltage source, it has a source voltage (also called the original voltage) $U_{\rm 0}$ and an internal resistance $R_{\rm i}$ . The internal resistance of a voltage divider considered as a voltage source results from the parallel connection of the divider resistors $R_{\rm 1}$ and $R_{\rm 2}$ :

$R_i = R_1 || R_2 = \frac{R_1\cdot R_2}{R_1+R_2}$

Use the measured values of resistors $R_{\rm 1}$ and $R_{\rm 2}$ to calculate the internal resistance $R_{\rm i}$ of the voltage source:

$R_i =$

$U_0 =$

The power $P_{\rm 0}$ supplied by the power supply can be calculated using the following equation:

$P_0 = U\cdot I_1$

The power consumed by the load resistance can be determined using the following formula:

$P_L = R_L\cdot {I_2}^2$

Fig. 1: Voltage divider

Draw the equivalent voltage source of the voltage divider:

What would be the value of $U_{\rm 2}$ without $R_{\rm L}$ ?

$U_{2, zero} =$

Calculate $U_{\rm 2,L}$ and $I_{\rm 2}$ for $R_{\rm L}$ = ${\rm 10} ~{\rm k\Omega}$ using the values of the equivalent voltage source: (Provide formulas!)

$U_{2L} :$

$I_2 :$

Check the values by measuring:

$U_{2L, Meas} :$

$I_{2, Meas} :$

Check the values using Kirchhoff's rules: (Provide formulas!)

$U_{2L} :$

$I_2 :$