Voltage divider as voltage source

The voltage divider shown in figure 5 is initially in the unloaded condition, because the entire current supplied by the power supply flows through the series-connected resistors $R_{\rm 1}$ and $R_{\rm 2}$. A resistor connected in parallel with $R_{\rm 2}$ loads the voltage divider.

Set the voltage on the power supply to $12 ~{\rm V}$ and measure the exact voltage with a multimeter.

For the connected load $R_{\rm L} = 10 ~{\rm k}\Omega$, the voltage divider represents a voltage source. Like any voltage source, it has a source voltage $U_0$ and an internal resistance $R_{\rm i}$.

The internal resistance of the voltage divider, considered as a voltage source, results from the parallel combination of the divider resistors $R_{\rm 1}$ and $R_{\rm 2}$:

$R_{\rm i} = R_{\rm 1} \parallel R_{\rm 2} = \frac{R_{\rm 1} \cdot R_{\rm 2}}{R_{\rm 1} + R_{\rm 2}}$

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 and determine the source voltage:

$R_{\rm i} =$


$U_0 =$ <

The power $P_0$ supplied by the power supply can be calculated using:

$P_0 = U \cdot I_{\rm 1}$

The power consumed by the load resistor can be determined using:

$P_{\rm L} = R_{\rm L} \cdot I_{\rm 2}^2$

Fig. 5: Loaded voltage divider

Draw the equivalent voltage source of the voltage divider.

What value would $U_{\rm 2}$ have without $R_{\rm L}$?

$U_{\rm 2,0} =$

Calculate $U_{\rm 2L}$ and $I_{\rm 2}$ for $R_{\rm L} = 10 ~{\rm k}\Omega$ using the values of the equivalent voltage source. State the formulae used.

$U_{\rm 2L}:$

$I_{\rm 2}:$

Verify the values by measurement.

$U_{\rm 2L,meas}:$

$I_{\rm 2,meas}:$

Verify the values using Kirchhoff's laws. State the formulae used.

$U_{\rm 2L}:$

$I_{\rm 2}:$