Question

The figure shows a circuit that illustrates the concept of loops, which are colored red and labeled loop 1 and loop 2. Loop 1 is the loop around the entire circuit, whereas loop 2 is the smaller loop on the right. To apply the loop rule you would add the voltage changes of all circuit elements around the chosen loop. The figure contains two junctions (where three or more wires meet)--they are at the ends of the resistor labeled R₃. The battery supplies a constant voltage Vb, and the resistors are labeled with their resistances. The ammeters are ideal meters that read I₁ and I₂ respectively.

The direction of each loop and the direction of each current arrow that you draw on your own circuits are arbitrary. Just assign voltage drops consistently and sum both voltage drops and currents algebraically and you will get correct equations. If the actual current is in the opposite direction from your current arrow, your answer for that current will be negative. The direction of any loop is even less important: The equation obtained from a counterclockwise loop is the same as that from a clockwise loop except for a negative sign in front of every term (i.e., an inconsequential change in overall sign of the equation because it equals zero).

Part B

Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R₂).

Answer in terms of given quantities, together with the meter readings I₁  and I₂  and the current I₃ .

Part C

Apply the loop rule to loop 2 (the smaller loop on the right). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal.

Express the voltage drops in terms of Vb, I₂, I₃, the given resistances, and any other given quantities.

Part D

Now apply the loop rule to loop 1 (the larger loop spanning the entire circuit). Sum the voltage changes across each circuit element around this loop going in the direction of the arrow.

Express the voltage drops in terms of Vb, I1, I₃the given resistances, and any other given quantities.

Answer 1

Part B)

Note that at junction 1, I2 and I3 are going "into" it whileI1 is "comingout". Assign a positive sign to currents flowing in and anegative to the ones flowing out. If you add all the currents you should geta zero: I2 + I3 - I1 = 0.

Part C)

Starting from point 1 and following the counter clockwise direction loop in red:

Since we are moving against I3, we get a "rise" in potential through resistor R3: I3R3

Through R2 we have a "drop" since we are moving in the same direction as I2: -I2R2

In a loop the sum of potentials differences must be zero, so: I3R3 - I2R2 = 0

Part D)

Staring from point 1 and using the same principles used in part C, but know moving clockwise

we get: Vb - I1R1 - I3R3 = 0