The question is incomplete. The complete question is :
The open circuit C containing ideal sources and resistors is measured at [tex]$V_{OC} = 10 \ V$[/tex] while the current through the short circuit passing through the terminals is [tex]$I_{SC} =100 \ mA$[/tex] . Find the operating point (V, I) that will occur at the terminals when circuit C is connected to circuit [tex]$C_2$[/tex] consisting of [tex]$V_2=15 \ V$[/tex] in parallel with [tex]$R_2=200 \ \Omega$[/tex] .
Solution :
Given :
[tex]$V_{OC} = 10 \ V$[/tex]
[tex]$I_{SC} =100 \ mA$[/tex]
[tex]$= 100 \times 10^{-3} \ A$[/tex]
[tex]$R_{th} =\frac{V_{OC}}{I_{SC}}$[/tex]
[tex]$=\frac{10}{0.1}$[/tex]
= 100 Ω
Therefore, by KVL, we get
[tex]$I=\frac{V_{OC}-V_2}{1/R_{th}+1/R_2}$[/tex]
[tex]$=\frac{10-15}{400/5}$[/tex]
[tex]$=-5 \times \frac{5}{400}$[/tex]
= [tex]$-\frac{1}{16}$[/tex] mA
By the KVL in loop,
-10 + 100 I + V = 0
V = 10 - 0.2
= 9.8 volts
