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Answers to Network Problem

• Since R1 and R2 are in series, and the current in R1 is 0.5A, the current in R2 is 0.5A.

• Using Ohm's law, V = IR, the voltage drop in R2 is V2 = (0.5A)(15ohms) = 7.5V.

• Since the voltage drop across R4 is 10V, R1 and R2 must share 10V and R3 must have a voltage drop of 10V, since these parts of the circuit are in parallel.  Then R1 = 10V - 7.5V = 2.5V.

• Using Ohm's law, R1 = V1/I1 = 2.5V/0.5A = 5ohms.  Also, R3 = V3/R3 = 10V/0.5A = 20ohms.

• The circuit current is 2A.  This is found by looking at R5.  Since R5 is connected to the battery in series, it has the full circuit current.  The current splits 3 ways in the top section.  The R1 and R2 section takes 0.5A and the R3 section takes 0.5A.  This means that the R4 branch must take 1A.  (0.5A + 0.5A + 1A = 2A, the circuit current)

• Using Ohm's law, R4 = V4/I4 = 10V/1A = 10 ohms.

• The voltage supplied by the battery is 50V.  The top section takes 10V and the bottom section takes 20V, this means that R5 has a voltage drop of 20V.  The 3 sections of this circuit are in series with each other, therefore each must take part of the voltage supplied.

• Using Ohm's law, R5 = V5/I5 = 20V/2A = 10ohms.

• The current in R7 is the same as the current in R6 since they are connected in series.  Therefore I7 = 1A.

• Using Ohm's law, V7 = (I7)(R7) = (1A)(7ohms) = 7V.

• R6 and R7 must share 20V since they are in parallel with R8.  Therefore, V6 = 20V - V7 = 20V - 7V = 13V.

• Using Ohm's law, R6 = V6/I6 = 13V/1A = 13ohms.

• Since the circuit current is 2A and the current splits 2 ways in the bottom part of the circuit, the current in R8 must be 1A.  (R6 and R7 take 1A, leaving 1A for the other part of this section.

• Using Ohm's law, R8 = V8/I8 = 20V/1A = 20ohms