A Network is a system of resistors and power sources connected in series and parallel. The key to solving these is to try and decide what the overall circuit equivalence is. For example:

Start by reducing the network to a single equivalent resistor. There are two 4 W resistors in parallel, and those are in series with an 8W resistor. The equivalent parallel resistance = 2 W (See Resistors in Series and Parallel) . This adds to the 8 W resistor for a total Req of 10 W. What is Itot?

I = V/Req = 20V/10W = 2Amps.

Now, the voltage drop across the 8 W resistor = IR = (2A)(8W) = 16 Volts. The voltage drop across the two 4W resistors must be 20V - 16V = 4V. This is what the voltmeter across that part of the network should read.

From Ohm's Law, I1 = V/R1 and I2=V/R2. So I1= 4V/4 W = 1Amp and I2 = 4V/4W = 1Amp. This makes sense, since Itot = I1 + I2.

Gustav Kirchoff provided 2 laws in the 19th century to help solve these networks.

  • Law 1: The sum of all currents entering a junction point must equal the sum of all currents leaving a junction point.
  • Law 2: The sum of the EMFs (voltage drops) across all the elements around any closed loop equals zero.

Now, consider the following conventions for a loop:

1) We will arbitrarily circle the loop in a clockwise direction.

2) If a resistor is crossed in the direction of current, the potential across that resistor is -IR.

3) If a resistor is crossed opposite the current, the potential across that resistor is +IR.

4) If a voltage source is crossed from negative to positive, it is +V.

5) If a voltage source is crossed from positive to negative, it is -V.

Let's look at a more complex example:

From Kirchoff's First Law, I1 + I2 = I3. We will move around the loops in a clockwise direction and see if we have chosen our signs correctly. Keep our sign conventions from above in mind. Add the potential drops for each loop (abef and bcde).

Loop bcde: 6V - (4 W)(I1) + (6 W)(I2)- (2 W)I1 = 0 or 6I1 - 6I2 = 6V

Loop abef: 12V -(6 W)(I2) -(2 W)(I3) = 0 or 6I2 + 2I3 = 12V

Substituting I1 + I2 = I3 , the loop abef equation becomes 2I1 + 8I2 = 12V

Now, add the two equations together to solve for I2. This gives us I2 = 1A. Since it is positive, we have chosen the direction of I1 correctly. Substituting this value back into the first equation yields I1 = 2 A. Again, since it is positive, we have chosen our direction correctly. Finally, substituting these two values into I1 + I2 = I3 gives I3 = 3A. Again, I3 is flowing in the direction shown.

For more on Networks and Basic DC Theory, try:

The NTNU (National Taiwan Normal University) Virtual Physics Laboratory provides excellent resources and I'm a big fan. Click here to read up on Ammeters, Voltmeters, and Multimeters.

For Practice Problems, Try:

Giancoli Multiple Choice Practice Questions (Go ahead - try a few.)