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The topic of this problem is Kirchhoff's Voltage Law.

Â The problem is, to write the mesh or

Â loop equation associated with this single loop circuit.

Â So in order to do that, the first thing we have to do is we have to assign

Â a current direction so that we can then assign polarities

Â to the voltage drops across the resistors, using the passive sign convention.

Â The passive sign convention was mentioned earlier in videos in chapter one and

Â two, so if you need to get familiar with

Â information related to the passive sign convention, you can look at those videos.

Â So the first thing we do is we assign a current, and we're going to assign

Â the current in a clockwise Direction around our single loop circuit.

Â So if we do that, we know that through this passive sign convention,

Â the current should enter the positive end of the voltage drops across our resistors.

Â So we have these voltage drops assign these polarities across our resistors.

Â Or we have a VR2 for voltage across resistor 2,

Â VR1 across the resistor 1, and VR3 across resistor 3.

Â Now that we have done that, we can now write our mesh,

Â also known as loop equation, for this single loop circuit.

Â So if we start at the lower left-hand corner of our circuit and

Â we work our way around the circuit in a clockwise fashion, the first thing we hit

Â is a -30 volts associated with the independent voltage source.

Â We continue around our circuit and then we encounter the resistor.

Â And the resistor is a positive voltage drop, + VR1.

Â 1:48

Continue past that resistor onto the next independent voltage source and

Â it's -5 volts.

Â Continuing around the loop to R2,

Â we have another voltage drop, VR2.

Â Continuing around the bottom portion of our single loop circuit, we have

Â a -15 volts with a independent voltage source at the bottom of the circuit.

Â And then we have one more voltage drop across VR3, and that is equal to 0.

Â So that's our equation for this single loop circuit.

Â