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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