The topic of this problem is circuits with dependent sources and the problem is to determine the current in the circuit and the output on load voltage V sub 0. So if you look at the circuit that we’ve drawn, we have a 12 volt source at the left hand side of the circuit and we have our load resistance at the right hand side of the circuit. So we have a 5 kilo ohm load resistance on the right hand side of the circuit. We also notice that we have a dependent source. The dependent source is at the top of the circuit, and it's a current controlled voltage source where the voltage associated with that dependent source is controlled by the current that's flowing through this single loop circuit. So we're going to again determine the current and the output voltage associated with this circuit. So in order to do this, we want to use Kirchhoff's Voltage Law and we can do that by summing up the voltages as we go around the circuit and we could choose either a clockwise or a counterclockwise direction to add these voltages up, the voltage drop associated with each element. It really doesn't matter whether we go clockwise or counter clockwise. In this case, we're going to start with the lower left hand corner of the circuit and we're going to go clockwise around the circuit. We see that if we do that, the first thing we encounter is a minus 12 volts associated with the source. So again, we're doing Kirchhoff's Voltage Law and it's clockwise around the circuit. So it's minus 12 volts associated with the voltage source on the left hand side of the circuit and there we have the voltage drop associated with the 3 kilo ohm resistor and that's going to be I1 times 3 kilo ohms. And using the passive sign convention, we designate the voltage drop as plus the minus across the 3 kilo ohm source, 3 kilo ohm resistance sorry. So we continue going around and we run into the dependent source and for the dependent source, we have a negative voltage associated with that. Since we hit the negative polarity of the voltage source first, so it’s minus V sub A and we continue around. We also have using passive sign convention, a plus to minus voltage drop across a 5 kilo ohm load resistance. And so that is going to be plus I1 times 5 kilo ohms is equal to 0. So that's our loop equation using Kirchhoff's voltage law for this a single loop circuit. So minus 12 volts plus I1 times 3K, the voltage drop across the 3 kilo ohm resistor minus V sub A since we hit the negative polarity of the dependent source first, as we travel clockwise around the circuit, and then we have the voltage drop I sub 1 5K across the 5 kilo ohm resistance. And so that gives us an equation with Two unknowns I1 and V sub A. We also have our relationship so that's our first equation. We also know the relationship between V sub A and I1 simply through our dependent source and that gives us our second equation that V sub A is equal to 2000 which is the amplification factor times I1. So that's our second independent equation relating VA and I1. So if we use those two equations and we solve for I1, we end up with an I1 equal to 2 milli amps. And using that, we can then solve for our output voltage. Our output voltage is going to be I1 times 5 kilo ohms. And we've solved for I1 already, and so we end up with an output voltage which is equal to 10 volts.
