In general, circuits contain both parallel and series branches and we must be able to analyze them. An example of such a circuit appears below:
A complete analysis of this circuit would involve determining the voltage across and the current through each resistor. The general approach to this type of circuit analysis is to replace parallel and series combinations of resistors by their equivalent resistances, until the circuit has been simplified enough to apply Ohm's Law. The best way to see how this works is to look at some examples.
Example 1:
What is the current through, and the voltage across each resistor in the circuit below?
Solution:
We begin by noting that we could simplify the circuit to one with only two parallel branches, if we found the equivalent of series resistors R2 and R3. We have already learned that the equivalent resistance of resistors in series is:
Req = R1 + R2 + R3 .... + RN
Since there are only two resistors to worry about, the formula becomes:
Req = R2 + R3 = 390Ω + 270Ω = 660Ω.
We can now simplify our circuit as shown below:
Now we have a circuit with only parallel branches, which we learned how to analyze in the last section. We can apply Ohm's Law to each branch to get:
IR1 = V/R1 = 12V/330Ω = 0.0364A = 36.4mA
IReq = V/Req = 12V/660Ω = 0.0182A = 18.2mA
The total current flowing from the battery is the sum of the currents in each parallel branch:
ITOTAL = IR1 + IReq = 36.4mA + 18.2mA = 54.6mA
Since R2 and R3 are in series, the current that flows through Req is the same as the current flowing through R2 and R3. Thus:
IR2 = IR3 = IReq = 18.2mA
Now we know all the currents. We can determine the voltage across each resistor by looking at the circuit and applying Ohm's Law in a different form, V = IR, where necessary.
First of all, we see that the voltage across R1 is just the source voltage, 12V, because R1 is connected directly across the source.
Second, we can use the current through R2 and R3, together with Ohm's Law, to find the voltage across each resistor:
VR2 = (IR2)(R2) = (18.2mA)(390Ω) = 7.1V
VR3 = (IR3)(R3) = (18.2mA)(270Ω) = 4.9V
Here is a more complex example:
Example 2: Find the current flowing through R8
Solution:
We must simplify the circuit so that we can apply Ohm's Law. At good place to begin is with R7 and R8. They are in series and can be replaced by a single resistor as shown below. We can use our formula for resistors in series to compute the resistance of the resistor that replaces R7 and R8:
Req = R7 + R8 = 18K + 33K = 51K
We need to do some more simplification of the circuit. The 51K resistor is in parallel with R6, and we could replace these resistors by an equivalent resistor whose resistance is given by:
Req = (R6*51K)/(R6+51K) = (820*51000)/(820+51000) = 41,820,000/51,820 = 807 ohms.
Now we can replace the parallel combination of the 51K and 820 ohm resistors by a 807 ohm resistor, as shown below.
Next, we combine R3 and R5. These resistors are in series, so the equivalent resistance is just the sum of the individual resistances:
Req = R3 + R5 = 100 + 9.1K = 9.2K
This further simplifies the circuit, as shown below:
At this point, we can make yet another simplification, by replacing the parallel combination of the 9.2K resistor and R4 with its equivalent resistance:
Req = (9.2K*R4)/(9.2K + R4) = (9200*220)/(9200+220) = 2,024,000/9420 = 215 ohms.
The new equivalent circuit is shown below:
We have simplified the circuit considerably, but we are not at a point where we can apply Ohm's Law. Our next step is to combine the 215 and 807 ohm resistors. They are in series, so the equivalent resistance is just the sum, 1022 ohms. Now are circuit looks like this:
We are almost finished!! Now we replace the parallel combination of R2 and the 1022 ohm resistor with its equivalent resistance:
Req = (R2*680)/(R2 + 680) = (1022*680)/(1022+680) = 694,960/1702 = 408 ohms
Now we have the following circuit:
Our final step is finding the equivalent resistance of the series combination of R1 and the 408 ohm resistor. The equivalent resistance is just the sum, 878 ohms, and we now have the following equivalent circuit:
Now we can apply Ohm's Law to find the current through the resistor:
ITOTAL= V/R = 42V/878 ohms = 0.0478 A = 47.8 mA.
This current is the total current flowing from the voltage source into the network formed by resistors R1 through R8. We need to determine the current flowing through R8 alone. To do this, we work our way back through our sequence of equivalent circuits.
Consider the circuit below:
This is one of the earlier equivalent circuits we developed for our original circuit. R8 is part of the 408 ohm equivalent resistance, so we need to determine the voltage drop across it in order to eventually find the current through R8. That volage drop is given by Ohm's Law:
V408 = I 0.0478A*408 ohms = 19.52 V.
Now we work backwards one step farther and decompose the equivalent resistance of 408 ohms into a parallel combination of R2 and an equivalent resistance of 1022 ohms, as shown below:
Since R6 and the 1022 ohm resistor are in parallel, the voltage drop across each resistor is equal to the voltage drop across the parallel combination. R8 is contain in the 1022 ohm equivalent resistor, we will need to know the current that flows through the 1022 equivalent resistance. We can use Ohm's Law once again:
I = V/R = 19.52/1022 = .0191 A
We must now go another step, and decompose the 1022 ohm resistance back into an equivalent resistances of 807 ohms and 215 ohms in series:
The 215 ohm resistor is a combination of R3, R4, and R5, which we do not need to work with further. R8 is part of the 807 ohm equivalent resistance, so we need to know the voltage drop across the 807 ohm resistor:
V = IR or V807 = 0.191*807 = 15.41V
Next, we decompose the 807 ohm resistance into a parallel combination of R6 (820 ohms) and a 51K ohm resistor (which is a series combination of R7 and R8).
We know that the voltage across each element of the parallel combination is equal to the parallel combination, which makes the voltage across the 51K resistor equal to 15.41V. Since the 51K resistor is a series combination of R7 and R8, the current that flows through the 51K equivalent resistance is the same as the current flowing through R8. We are almost done! We can apply Ohm's Law once again to find the current:
I R8= V/R = 15.41V/51000 = 0.000302 A = .302mA = 302μA
That is the current flowing through R8 and we are done!! At this point, you may wonder if there is a better way to do this type of analysis and the answer is yes. In the next two sections we will two different techniques that can greatly simplify the analysis of complex circuits.
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