vs =6 volts

vth =2 volts
ida(diode current)=1m amp
no value given for resistor network
find rth(thevenien)
2,000 and 4000 do not work.
diode current must not be equal to isc.??
voc=2 volts

Diode Connected:

Vd + (I2+Id)R1 = 6 Volts.
0.7 + (I2+1.0)R1 = 6.
0.7 + I2*R1 + R1 = 6.
Eq1: I2R1 + R1 = 5.3.

Diode disconnected:
V2 + I2*R1 = 6 Volts.
2 + I2R1 = 6.
Eq2: I2R1 = 4.

In Eq1, replace I2H1 with 4:
4 + R1 = 5.3.
R1 = 1.3k = 1300 Ohms.

In Eq2, replace R1 with 1.3k:
I2*1.3k = 4.
I2 = 3.1 mA.
R2 = V2/I2 = 2/3.1 = 0.65k = 650 Ohms.

Rth = R1*R2(R1+R2)=1300*650/(1300+650)= 433 Ohms.



r

To find the Thevenin equivalent resistance (Rth), we need to use the given information of the voltage source (Vs), the Thevenin voltage (Vth), and the diode current (Ida). However, you mentioned that no value is given for the resistor network, which is required to calculate the Thevenin resistance accurately. Without the resistor network values, it is not possible to determine the Thevenin resistance (Rth).

Regarding your statement that 2,000 and 4,000 do not work, I assume you are referring to using those values for the resistor network. Again, without knowing the actual resistor values, it is difficult to determine which values would work.

Also, you mentioned that the diode current (Ida) must not be equal to the short-circuit current (Isc). Generally, this is correct. The short-circuit current (Isc) is the current that flows through the circuit when it is shorted, usually when the load is removed. The diode current (Ida) should be smaller than the short-circuit current (Isc) to ensure that the diode is not damaged or operated outside its safe limits.

Lastly, you mentioned the open-circuit voltage (Voc) is 2 volts. The open-circuit voltage is the voltage across the output terminals when there is no load connected. This information alone does not provide enough detail to determine the Thevenin resistance (Rth).