# Resistor Combination Exercise

Calculate the current flowing through each resistor in the following circuit:

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.

1/Reff = 1/R + 1/R = 2/R, so that
Reff = R/2

This value is then used to replace the pair of 240 ohm resistors with a single effective resistor of 120 ohm, as shown below.

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

Reff = R/2 + R, so that
Reff = 3/2 R

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

I = V / Reff = (120 V) / (360 ohm) = 0.33 A

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 V1 across a 120 ohm resistance with a current of 0.33 A is determined by Ohms's law: V1 = I Reff = (0.33 A)(120 ohm) = 40 V. Similarly V2 = 80 V across the 240 ohm resistor with a current of 0.33 A flowing. Note that the sum of voltages is 120 V, the value applied by the battery, as it must.

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:
R1: current 0.17 A, voltage 40 V
R2: current 0.17 A, voltage 40 V
R3: current 0.33 A, voltage 80 V