To understand the origins of both of Kirchhoff\'s rules and how to use them to s

Question

To understand the origins of both of Kirchhoff's rules and how to use them to solve a circuit problem.

This problem introduces Kirchhoff's two rules for circuits:

Kirchhoff's loop rule: The sum of the voltage changes across the circuit elements forming any closed loop is zero.

Kirchhoff's junction rule: The algebraic sum of the currents into (or out of) any junction in the circuit is zero.

The figure (Figure 1) 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 imporant: 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 A

The junction rule describes the conservation of which quantity? Note that this rule applies only to circuits that are in a steady state.

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.

Explanation / Answer

(A) Think of the analogy with water flow. If a certain current of water comes to a split in the pipe (mathematically) the sum of the three water currents at this junction has to be zero. If this were not true, water would accumulate at the junction. Same applies here to the current. In steady state, current that eneters a node must equal the current that leaves the node, hence its essentially conservation of current.

(B) The current through resistance R1 is not labeled. You should recognize that the current I1 passing through the ammeter also passes through resistance R1 because there is no junction in between the resistor and the ammeter that could allow it to go elsewhere. Similarly, the current passing through the battery must be I1 also. Circuit elements connected in a string like this are said to be in series and the same current must pass through each element.

Applying junction rule:

Current enetering 1 = Current leaving 1, i.e.

I2 + I3 = I1

or, I2 + I3 - I1 = 0

(C) In determining the signs, note that if your chosen loop traverses a particular resistor in the same direction as the current through that resistor, then the end it enters through will have a more positive potential than the end from which it exits by the amount IR. Thus the voltage change across that resistor will be negative. Conversely, if your chosen loop traverses the resistor in the opposite direction from its current arrow, the voltage changes across the resistor will be positive. Let these conventions govern your equations (i.e., don't try to figure out the direction of current flow when using the Kirchhoff loop--decide when you put the current arrows on the resistors and stick with that choice). Now, an ideal ammeter has zero resistance. Hence there is no voltage drop across it.

Expressing the voltage drops in terms of I2 and I3, for loop 2,

I3R3 - I2R2 = 0 = -I3R3 + I2R2

(D) Sum the voltage changes across each circuit element around this loop going in the direction of the arrow.

Vb - I1R1 - I3R3 = 0

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