• Example 0:

  

  The voltage measured by a voltmeter is (voltage divider):
  

  $$V=V_s \frac{R_v}{R_s+R_v}$$
  
  
  It is desireable for $R_v$ to be as large as possible (ideally infinity) so that it draws little current from the source (causing little voltage drop across the internal resistance $R_s$), for the measureed voltage to be close to the true open-circuit voltage $V_{oc}$, i.e., the source $V_s$.

  The current measured by an ammeter is:
  

  $$I=\frac{V_s}{R_s+R_a}$$
  
  
  It is desireable for $R_a$ to be as small as possible (ideally zero) so that the measured current is close to true short-circuit current $I_{sc}=V_s/R_s$.

  When $R_v<\infty$ and $R_a>0$, the true $V_s$ and $R_s$ can be found by solving these equations:
  

  $$\left\{ \begin{array}{l} V=V_s R_v/(R_s+R_v) \\ I=V_s/(R_s+R... ...ay}{l} R_vV_s-VR_s=VR_v \\ V_s-IR_s=I R_a \end{array} \right.$$
  
  
  solving these we get:
  
  $$\left\{ \begin{array}{l} V_s=IV(R_v-R_a)/(IR_v-V) \\ R_s=(V-IR_a)R_v/(IR_v-V) \end{array} \right.$$
  
  

  Given the specific values, we get
  

  $$I=\frac{R_s}{R_s+R_a}=\frac{6V}{300\Omega}=0.02A$$
  
  
  
  
  $$V=V_s \frac{R_v}{R_s+R_v}=6V \frac{10,000}{10,000+100}=5.94 V$$
  
  
  From these measurement, the true $R_s$ and $V_s$ can be obtained as:
  
  $$V_s=\frac{IV(R_v-R_a)}{IR_v-V} =\frac{5.94\times 0.02(10,000-200)}{0.02\times 10,000-5.94}=100$$
  
  
  
  
  $$R_s=\frac{(V-IR_a)R_v}{IR_v-V} =\frac{(5.94-0.02\tiimes 200)\times 10,000}{0.02\times 10,000-5.94}=6$$
  
  
  Method verified.