Example: Finding the Voltage and Current across each Resistor in a Complicated Network Example: Finding the Voltage and Current across each Resistor in a Complicated Network

Other examples:
Giancoli Example 19-7 (pg 528)

Finding the voltage over resistors in a complicated network is done using Ohm's Law, V=IR. Just as the battery voltage is the total current multiplied by the total equivalent resistance, so the voltage over any given resistor is the current through that resistor multiplied by its resistance. Consequently, to find the voltage over each resistor, you must first find the total current in the circuit, and then find the current through each branch. This can quickly become comlicated for a network with anything but trivial complexity. I will do the analysis for the following circuit:

1) Find the total resistance
First, I need to write each parallel element as one resistor:

Resistor network A:
1/Ra=1/20+1/15
Ra=8.57Ω

Resistor network B:
1/Rb=1/25+1/10
Rb=7.14Ω

Total Resistance:
Rt=Ra+Rb+20=35.71Ω

2) Calculate the total current
I'll say that all this is connected to a 1V battery to keep the numbers a little bit easier.
I=V/Rt=.028A

3) Calculate the voltage over Ra, Rb, etc.
Reducing the parallel network to equivalent resistors in series makes the network look like this:

The voltage across equivalent resistor Ra is Va=I*Ra, and similarly for the other series elements. Here I use the total current I calculated above because I'm dealing with a series circuit, so the total current is the same everywhere (Lab 1).

Va=I*Ra=.028A*8.57Ω=0.24V
Vb=I*Rb=.028A*7.14Ω=0.20V
Vc=I*Rc=.028A*20.0Ω=0.56V

As a check, the sum Va+Vb+Vc=1, which is the total voltage of the battery. This is exactly what should be the case, as a charge traveling all the way across the network must use all its voltage.

4) Find how the current splits within each network
Va is the change in voltage over the 15ohm resistor, and the series of the two 10ohm resistors. So the current that flows through the 15ohm resistor is I=V/R:
I(15Ω resistor)=0.24V/15Ω=.016A
The rest of the current (.028A-.016A=.012A) obviously must split off and go through the two 10ohm resistors. So the voltage drop across each is:
V(10Ω resistors)=.012A*10Ω=0.12V

As a quick check, each resistor takes 0.12V, so both together take 0.24V which is Va as it needs to be.

Vb is the change in voltage over the 10Ω resistor and 25Ω resistor in network B. So, the current through each resistor is:
I(10Ω resistor)=0.20V/10.0Ω=0.020A
I(25Ω resistor)=0.20V/25.0Ω=0.008A

Finally, the current through the lone 20Ω resistor is given by
I(20Ω resistor)=0.56V/20.0Ω=0.028A

Which is the total current, as it must be. To summarize:
Total Voltage: 1V
Total Current: 0.028A
15Ω Resistor (Network A): 0.24V, 0.016A
10Ω Resistors (Network A): 0.12V, 0.012A
25Ω Resistor (Network B): 0.20V, .008A
10Ω Resistor (Network B): 0.20V, .020A