Prior to completion of the previous lecture’s circuit problem, some additional practice identifying nodes and branches within a multi-loop circuit will be helpful:
We begin our journey at the 3 V source located at the far-left side of the diagram. As the current ( I ) moves upward and to the right, we encounter our first node. Please recall that the quantity of current that enters a node is equal to the quantity of current that emerges from it. A single wire enters the node ( a.k.a. “ junction “ or “ branch point “ ) whilst two wires emerge from it. If we label the current that enters the node as I1, the currents that emerge from it can be labeled I2 and I3:
I1 = I2 + I3
As I2 and I3 move along their designated branches, they encounter circuit elements. Current I2 passes straight downward through a single 75 ohm resistor, and I3 passes through a 12 V source and a 125 ohm resistor on the adjacent branch. Notice how the 12 V source is situated in such a way that its positive ( wide ) terminal opposes the flow of positive charge into it. Please be mindful that the direction of current flow has been arbitrarily chosen for convenience. Subsequent calculations will show whether or not the current path we’ve chosen mirrors the actual direction of current flow. Finally, currents I2 and I3 re-convene at a second node.
We must now determine how many electrical paths the circuit’s branches give rise to. At first, it may seem as if only two branches are present; however, we are at liberty to consider the outermost perimeter of the circuit to be a branch if needed. For the time-being, it is most convenient to consider our system as being comprised of two loops:
We are now ready to apply Kirchhoff’s Loop Rule to each loop:
We begin my moving upward and across the 3 V source and around Loop 1. As we move around this loop, we must be sure to use the appropriate current designations as we cross the resistors in the circuit. For the sake of simplicity, we will now omit the voltage suffix from our calculations:
Loop 1:
3.0 – I2R1 = 0
3.0 = I2R1
I2 = ( 3.0 / R1 )
I2 = ( 3.0 / 75 Ω ) = 0.04 A
Loop 2:
– 12 – I3R2 + I2R1 = 0
Note: We determined early on that the I2 current travels straight downward. Traveling with the current within Loop 1 gives us a voltage drop across R1; thus, we have a rise in voltage as we travel back upward and across R1 to complete Loop 2.
– 12 – I3R2 + [ ( 0.04 A )( 75 Ω ) ] = 0
– 12 – I3R2 + 3 = 0
– 9 – I3R2 = 0
– 9 = I3R2
I3 = ( – 9 / 125 Ω ) = – 0.072 A
Since I3 has a negative value as calculated, its direction-of-travel within the diagram must be reversed:
With the values of I2 and I3 having been determined, we may now solve for I1 :
I1 = I2 + I3
I1 = 0.04 A + ( – 0.072 A )
I1 = – 0.032 A
We must now reverse direction of current I1 within our diagram:
Check:
Loop 1: – 3V + 3V = 0 V
Loop 2: 12V – 3V – 9V = 0 V
RENAISSANCE PHYSICS 