Series-Parallel Resistive Circuits

I. Basics of Series-Parallel Resistive Circuits

                                                               

                            . . .     . . . . . .

                                                               

Combination resistive circuits, otherwise known as Series-Parallel resistive circuits combine resistors in series with resistors in parallel as shown in the Figure above.  This kind of circuit can be separated into two different cases: For the part(s) of the circuit that has resistors in series we use our knowledge of series circuits and apply the corresponding principles, for the part(s) of the circuit that has resistors in parallel we use our knowledge of parallel circuits and apply the corresponding principles. 

Kirchoff's Voltage Law (KVL) is extended to include any loop on the circuit such that:  "The voltage drops around any closed loop in the circuit add up to zero."   Also, Kirchhoff's Current Law (KCL) is  extended to include any node on the circuit such that:  "The  currents coming into any node in the circuit add up to zero."

The equivalent resistance of the circuit Req can be obtained by making the appropriate combinations (either parallel or series)  until we get one unique resistance.  The rule that Vtotal = I*Req still applies as well as Ohm's Law and both should be kept in mind when solving combination circuits.

 

 

II. Some examples of finding Req

 

       Example 1

 

 

In the circuit above there are four resistors and a battery V=10 volts.  To obtain Req for this circuit notice

that R2 and R3 are in parallel, combining these two resistors R2||R3 = 1/(1/2 + 1/10) where the default unit is

Kilo-Ohms.  Obtaining the common denominator and combining gives R2||R3 = 1/(6/10) = 10/6 KOhms 

or R2||R3 = 1.67 KOhms.   The value of 1.67 KOhms will then be in series with R4.  Combining R4

with  the 1.67KOhms gives 6.67KOhms.  Finally this value of  6.67KOhms will be in parallel with R1.

Combining R1 with the 6.67KOhms results in  Req =  0.869KΩ.

Of course, once Req is found the total current I can be calculated by: I = V/Req = 10 volts/0.869KOhms,

which gives  I = 11.5 mA.


 

       Example 2

 

 

For this example there is a total of 5 resistors and for the sake of convenience the battery voltage is not

 

specified, since the calculation is only for the equivalent resistance Req.  This type of circuit is called a ladder

 

circuit and the way to handle this problem is by starting from the end and going towards the front.

 

Hence, first R5 and R4 can be combined in series to give a total value of 2KOhms.  The R4-R5 combination

 

can then be combined with R3 in parallel, such that  R3|| 2K = 2K||2K= 1KOhm.  Furthermore, this result

 

can be combined in series with R2 to give R2 + 1K= 2K + 1K = 3KOhms.  Finally, Req is the parallel

 

combination of the 3KOhm resistance with R1 which gives Req = R1||3K = 1K||3K = 1/(4/3) = 0.75KΩ.

 


 

            Example 3

 

 

 

One more example will suffice to illustrate how to obtain Req.  For this problem it is not necessary to start at

 

the end.  There is three different options here: R7 is in series with R8, R5 is in parallel with R6 and R3 is in

 

series with R2.  Starting with R2 and R3 they combine to give 2KOhms.  On the other hand R5 in parallel with

 

R6 gives 4KOhms, and R7 in series with R8 comes out to 4KOhms.  Now, to keep track of all these values

 

it is a good idea to redraw the circuit at this stage.   This is shown in the Figure below.

 

Notice that the circuit looks as Example 2, hence resistances can be combined starting from the end and

 

going to the front.  The two 4K resistors in parallel give a 2KOhm which can then be combined with R4

 

in series to give 4KOhms.  Then, this 4KOhm resistance is combined in parallel with the 2KOhm right at

 

the front end to give a value of  1.33KOhms.  Finally, this 1.33KOhms in series with R1 comes to a total

 

Req = 2.33KOhms.

 


III. Applying Kirchhoff's laws and Ohm's law in combination circuits.

                       

           I1 = I2 + I3                                                       V1 = V2 + V3

 

 

Example 1

For this first example the battery voltage in the circuit is 5 V and all the resistors have a value of 2KOhms,

 

except for R2 which is only 1KOhm.  The task at hand is to find all the voltages and currents in the circuit

 

as well as the equivalent resistance Req.  A good starting point is to find Req, since it will be used to find

 

other parameters.  By inspection, R4 and R5 in series give 4KOhms, which if combined with R3 results in

 

1.33KOhms.  The 1.33KOhm value can be combined with R2 in parallel to give a value of  0.57KOhms.

 

Finally, Req = 2K + 0.57K = 2.57KOhms.

 

Applying KCL to the circuit the total current coming from the battery  I1 goes through resistor R1 and

 

splits into I2 which goes through resistor R2, I3 which goes through R3 and I4 which goes through R4

 

and R5.  Hence KCL : I1 = I2 + I3 + I4. 

 

Since the current I1 is the total current in the circuit, applying Ohm's law this current is I1 = Vtotal/Req

 

which gives  I1 = 5V/2.57K =  1.95mA.   The voltage across R1, V1 = I1 * R1 = 1.95mA* 2K= 3.9V.

 

Applying KVL on the first loop, V2 (the voltage across R2) and the voltage V1 have to add up to be 5V.

 

Thus, V2 = 5V - V1 = 5V - 3.9V = 1.1V.    Since R2 and R3 are in parallel this same voltage is seen

 

through resistor R3, making V3 = V2 = 1.1V.

 

Going back to Ohm's Law once more, I2 and I3 can be easily found since we already have the voltage

 

across both resistors and know the resistor values.  Hence, I2 = V2/R2 = 1.1V/1K = 1.1mA , and

 

similarly, I3 = V3/R3 = 1.1V/2K = 0.55mA.

 

Finally, the current I4 can be found by applying KCL and noting that I4 = I1 - I2 - I3.  This equation gives

 

I4 = 1.95mA - 1.1mA - 0.55mA = 0.3mA.   Since R4 and R5 have the same 2KOhm value, both V4 and

 

V5 are equal, giving that V4 = V5 = I4* 2KOhms = 0.3mA * 2K = 0.6V

 


 

Example 2

 

One more example shall suffice to explain how to solve combination circuits.  Example 2 has two 1KOhm

 

resistors in series with 3KOhm resistors in parallel.  Again, it is a good idea to start out by finding Req.

 

Combining the 3K resistors gives a resistance of 1KOhm.  This 1KOhms resistance in series with R1

 

and R2 results in  Req = 3KOhms. 

 

The next step is to find all the currents and voltages involved in the circuit.  The current  I1 = I2 will be the

 

total voltage over Req;  hence:  I1 = I2 = 5V/3K = 1.67mA.   The voltages V1 and V2 can be calculated

 

by Ohms Law, since both resistors are the same value and carry the same current  then V1 = V2.

 

This gives V1 = V2 = 1K * 1.67mA = 1.67V.

By KVL the voltage across the 3KOhm resistors plus V1 plus V2 has to be equal to V = 5V.  From this,

it follows that the voltage across the 3KOhm resistors is V(3K) = 5V - 1.67V - 1.67V = 1.66V.

Also, V3 = V4 = V5 = 1.66V.    Finally, the currents through any of the 3KOhm resistors can be easily

obtained by Ohm's Law, so that  I3 = I4 = I5 = 1.66V/3KOhms = 0.553mA.    As a simple KCL check

the results can be confirmed by adding I3 + I4 + I5 which results in I2 or I1!

 

 A Note on Power:

 

          The total power in a combination circuit is by no surprise the sum of the power through

          the individual components, such that Ptotal = P1 + P2 + P3 + ... + Pn, where n is the

          number of components (resistors) in the circuit.  For parallel resistors the voltage is

          common to all components, so that P = V2/Rx, where Rx is the individual resistor value.

          For series resistor the current is common and  P = I2*Rx, Rx having the same meaning.

 

 

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