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Recall from last lecture:

18.3 Resistors in Parallel (cont'd)

Here's a much more difficult example with a circuit containing resistors in series and parallel.

Example: P18.13

The resistance between terminals a and b in Figure P18.13 is 75W. If the resistors labeled R have the same value, determine R.

(Draw diagram.) The resistors R are part of the resistor combination, so if we write out the expression for the equivalent resistance, then we can relate it to R and solve for R. This circuit is a combination of resistors in series and parallel, proceed step by step to find the equivalent resistance. We want to find the "deepest nesting" of resistors, and beginning there, work our way out to the whole circuit.

Begin by combining the series combination of R and the 5.0W resistor, Req1 = R + 5.0W. Next we can find the equivalent resistor for the parallel combination of the 120, 40, and Req1 resistors:
1/Req2 = 1/120 + 1/40 + 1/(R+5) = (1+3)/120 + 1/(R+5) = 1/30 + 1/(R+5) = (R + 5 + 30)/(30(R+5)) = (R+35)/(30(R+5))
Req2 = (30R+150)/(R+35)
Finally, we sum that equivalent resistance with the second resistor, R, in series with it (the one next to terminal a), and set the result equal to 75W:
Req = 75W = R + Req2 = R + (30R+150)/(R+35) = (R² + 35R + 30R + 150)/(R + 35) = (R² + 65R + 150)/(R + 35)
Therefore, 75 = (R² + 65R + 150)/(R + 35). To find R, multiply both sides by (R + 35):
75(R + 35) = 75R +2625 = R² + 65R + 150
R² - 10R - 2475 = 0
This quadratic equation has two possible solutions:
R = [10 ± Sqrt{10² + 4(2475)}]/2 = [10 ± Sqrt{10000}]/2 = [10 ± 100]/2 = 5 ± 50 = 55W or -45W
Resistances are positive, therefore the only logical answer is R = 55W.

18.4 Kirchhoff's Rules and Complex DC Circuits

With the results for combining pairs of resistors we can analyze nearly any circuit made from combinations of resistors and one voltage source. (The exceptions are circuits like the "current balance".) To analyze circuits that contain more than one voltage source, or both resistors and capacitors, we need Kirchhoff's rules. There are two rules:

  1. The sum of the currents entering any junction must equal the sum of the currents leaving.
  2. The sum of the potential differences (voltages gains and drops) across all the elements around any closed loop of the circuit must be zero.
A junction is any point in the circuit where three (or more) lines meet. (Drawing of a simple circuit with a junction.) The conductor must remain neutral (uncharged). Where three conductors meet to form a junction this must still hold, and that is possible only if the amount of charge entering the junction is balanced by an equal amount leaving the junction. Since current is just the charge moving past a location on the wire per unit time, by dividing the amount of charge entering or leaving the junction by the time interval of measurement, we get Kirchhoff's first rule. The first rule is equivalent to conservation of charge, and is known as the junction rule.

A closed loop of the circuit is any path through the circuit that begins and ends at the same point. (Indicate closed loops on the circuit drawing.) A charge moving around the loop will gain and lose energy as it passes through circuit elements, but it must return to its starting point with zero net change of energy. The second rule is equivalent to conservation of energy, and is known as the loop rule.

There is a well defined procedure for applying Kirchhoff's rules to a circuit.

  1. Assign names and directions to the currents in each branch of the circuit. Don't worry if you happen to choose the direction opposite the correct choice, you will simply get a negative value for the current. The magnitude will be correct, and the negative sign means that the actual direction is opposite to your initial assignment.
  2. Choose the loops, and the direction to traverse each loop. Record the voltage gain or drop for each circuit element traversed, according to the following rules:
    1. If a resistor is traversed in the direction of the current, then the voltage change is -IR.
    2. If a resistor is traversed in the direction opposite the current, then the voltage change is +IR.
    3. If a voltage source is traversed from - to +, then the voltage change is +V.
    4. If a voltage source is traversed from + to -, then the voltage change is -V.

For a complete loop, the sum of the voltage changes is zero. Now the loop equations and junction equations can be combined to eliminate unknowns until one quantity can be determined. Then using that quantity, others can be determined until all the unknowns are solved for. This is a complete solution of the circuit.

How many loop and junction equations are required for a complete solution of a given circuit? You need as many loop equations as there are "holes" in the circuit. You need one fewer junction equations. For example, the circuit in Figure 18.11 of the text has two holes, so two loop and one junction equations are needed. The circuit in Figure 18.40 has three holes, so three loop and two junction equations are needed.

Example: P18.16

Figure P18.16 shows a circuit diagram. Determine (a) the current, (b) the potential of wire A relative to ground, and (c) the voltage drop across the 1500W resistor.

This circuit has one hole, so we will need one loop equation and zero junction equations. (The green symbol in the bottom left corner is a ground symbol and indicates that that point in the circuit is at ground potential, or 0V.) Let's label the one current I, and define the current direction and the direction around the loop to be clockwise.

(a) Because we are always traversing the resistors in the direction of the current, the voltage change across each will be -IR. The 20V and 25V batteries are traversed from - to +, so their voltages will enter with a positive sign, but the 30V battery is traversed from + to -, so its voltage will be given a negative sign. The resulting loop equation is, starting from the lower left corner:
20.0V -I(2000W) -30.0V - I(1000W) - I(1500W) + 25.0V - I(500W) = 0
This can be simplified to:
15.0V -I(5000W) = 0    or
I = (15.0V)/(5000W) = 0.00300A = 3.00mA

(b) The potential of wire A relative to ground is given by the total change in voltage when a charge moves from ground to A. Moving up the left side of the circuit from ground to A, we gain 20.0V across the first battery, then lose -IR = -(3.00mA)(2000W) = -6.0V across the first resistor, then lose -30.0V across the second battery, then lose -IR = -(3.00mA)(1000W) = -3.0V across the second resistor, for a net change of V = 20.0 - 6.0 - 30.0 - 3.0 = -19.0V.

The voltage drop across the 1500W resistor is IR = (3.00mA)(1500W) = 4.5V. The sign of the voltage drop can only be given if the direction is specified.

© Robert Harr 2000