Answer:
Incomplete questions check attachment for circuit diagram.
Explanation:
We are going to use superposition
So, we will first open circuit the current source and find the voltage Voc.
So, check attachment for open circuit diagram.
From the diagram
We notice that R3 is in series with R4, so its equivalent is given below
Req(3-4) = R3 + R4
R(34) = 20+40 = 60 kΩ
Notice that R2 is parallel to the equivalent of R3 and R4, then, the equivalent of all this three resistor is
Req(2-3-4) = R2•R(34)/(R2+R(34))
R(234) = (100×60)/(100+60)
R(234) = 37.5 kΩ
We notice that R1 and R(234) are in series, then, we can apply voltage divider rule to find voltage in R(234)
Therefore
V(234) = R(234) / [R1 + R(234)] × V
V(234) = 37.5/(25+37.5) × 100
V(234) = 37.5/62.5 × 100
V(234) = 60V.
Note, this is the voltage in resistor R2, R3 and R4.
Note that, R2 is parallel to R3 and R4. Parallel resistor have the same voltage, then voltage across R2 equals voltage across R34
V(34) = 60V.
Now, we also know that R3 and R4 are in series,
So we can know the voltage across R4 which is the Voc we are looking for.
Using voltage divider
V4 = Voc = R4/(R4 + R(34)) × V(34)
Voc = 40/(40+60) × 60
Voc = 24V
This is the open circuit Voltage
Now, finding the short circuit voltage when we short circuit the voltage source
Check attachment for circuit diagram.
From the circuit we notice that R1 and R2 are in parallel, so it's equivalent becomes
Req(1-2) = R1•R2/(R1+R2)
R(12) = 25×100/(25+100)
R(12) = 20 kΩ
We also notice that the equivalent of Resistor R1 and R2 is in series to R3. Then, the equivalent resistance of the three resistor is
Req(1-2-3) = R(12) + R(3)
R(123) = 20 + 20
R(123) = 40 kΩ
We notice that, the equivalent resistance of the resistor R1, R2, and R3 is in series to resistor R4.
So using current divider rule to find the current in resistor R4.
I(4) = R(123) / [R4+R(123)] × I
I(4) = 40/(40+40) × 8
I(4) = 4mA
Then, using ohms law, we can find the voltage across the resistor 4 and the voltage is the required Voc
V = IR
V4 = Voc = I4 × R4
Voc = 4×10^-3 × 40×10^3
Voc = 160V
Then, the sum of the short circuit voltage and the open circuit voltage will give the required Voc
Voc = Voc(open circuit) + Voc(short circuit)
Voc = 24 + 160
Voc = 184V.
