a) Find v_o using source transforms.

We know that ANY circuit comprising sources and resistors has a Norton equivalent, so let's find the Norton equivalent of this current source in series with a resistor. We can then replace these two elements with the Norton equivalent.

First, find the short circuit current, I_sc. This will be the value of the Norton current source.

If it is not obvious that the short circuit current is - 1 amp, there is little hope for you unless you learn the basics VERY fast.

I_sc = - 1 A

Finding the open circuit voltage turns out to be a little problematic, so let's determine the Thevenin equivalent resistance directly. This may be done by "killing" the independent source and finding the equivalent resistance of the circuit that remains connected to terminals x and y. To deactivate a current source, we replace it with an open circuit (thus guaranteeing that the current is zero).

Behold! There is now an open circuit (or infinite resistance) between terminals x and y, thus the equivalent resistance is infinite (or an open circuit).

R_th = (open circuit)

Thus placing a - 1 A source in parallel with an open circuit gives the following equivalent connected to terminals x and y.

In other words, we can simply throw the 16 ohm resistor away (replacing it with a wire), and the effect on the rest of the circuit will be unchanged.

(NOTE: for those who are concerned about the voltage across the 16 ohm resistor no longer being there, remember that the voltage across a current source will be whatever it must to maintain the stated current. In this case, the voltage across the current source will change by 16 volts depending on whether the 16 ohm resistor is there or not.)

We know that ANY circuit comprising sources and resistors has a Thevenin equivalent, so let's find the Thevenin equivalent of this voltage source in series with a resistor. We can then replace these two elements with the Thevenin equivalent.

First, find the open circuit voltage, V_oc. This will be the value of the Thevenin voltage source.

Once again, if it is not obvious that the open circuit voltage equals 520 volts, you'd better do some basic learning really quickly.

V_oc = 520 V

Finding the short circuit current is a bit problematic, so let's find the equivalent resistance directly by "killing" the source and finding the equivalent resistance that remains. To deactivate a voltage source, we replace it with a wire (thus assuring that the voltage from z to x is zero).

Hopefully it is obvious that the resistance connected to x and z is zero (a resistor in parallel with zero ohms - work it out if you don't see this), so

R_Th = 0 W

and the Thevenin equivalent connected to terminals x and z is (the voltage source in series with zero ohms)

Thus we can just throw the 260 ohm resistor away.

(NOTE: for those who are concerned about the current through the 260 ohm resistor no longer being there, remember that the current through a voltage source will be whatever it must to maintain the stated voltage. In this case, the current through the voltage source will change by 2 amps depending on whether the 260 ohm resistor is there or not.)

Now we can apply a source transform.

If we transform the current source in series with the 40 ohm resistor connected to terminals x and y, that will place a voltage source and resistor in series with the other voltage source and resistors.

Combining the voltage sources in series and the three resistors to the LEFT side of v_o gives

Finally, voltage division will give the value for v_o.

v_o = 480 (250) / 300
v_o = 400 V