**A Solution:**

The approach to analyzing circuits in SCEN103 is to reduce the circuit to the equivalent simple circuit. This requires identifying series or parallel configurations pair by pair and reducing to one effective resistance. The two resistors in parallel can be reduced to an effective resistance that is one-half of the individual values.

Next, reduce the 2 "resistors" in series with an effective value.

Finally, Ohm's law may be applied to find the current flowing through the battery:

To find the current in each resistor and the voltage across each resistor, Ohm's law is applied systematically while "undoing" the effective circuit. For example, consider the intermediate circuit again:

The voltage **V _{1}** across a 120 ohm resistance with a current of
0.33 A is determined by Ohms's law:

Finally, the effective resistance of 120 ohm was formed from two 240 ohm resistors in parallel. We already know that each resistor experiences 40 V across it; from Ohm's law the current in each resistor is (40 V) / (240 ohm) = 0.17 A.

Note that the current of 0.33 A through the battery splits equally between the two paths of equal resistance.

In summary:

**R _{1}**: current 0.17 A, voltage 40 V

Back to SCEN103 Home Page.

Comments, suggestions, or requests to ghw@udel.edu.

"http://www.physics.udel.edu/~watson/scen103/exer3.html"

Last updated Sept. 17, 1996.

Copyright George Watson, Univ. of Delaware, 1996.