Parallel Resistive Circuits

I. Basics of Parallel Resistive Circuits

                       I1   -->          R1              In the Figure on the left we see several resistors in

                       I2   -->         R2              parallel.  If these are all attached to one another they

                       I3   -->         R3              will share the same voltage V.  However the current coming

                       I4   -->         R4             into the same node on the left (meaning the same contact

                                                                          point) will be split into several different currents : I1, I2

                                             :              I3, I4 ... In,  each one going through a separate resistor

           

                        In  -->         Rn             R1, R2, R3, R4 ... Rn  respectively.

 

                                           + V -

 

Our first principle for parallel circuits is that all resistors will share the same potential V, or:

 

1. The potential V is the same for all resistors.

 

The current coming into the left node (that is the left contact where all resistors are attached to) will

split such that the total current is the sum of the currents through all of the resistors- as a tree and its branches.

Hence the second principle for parallel circuits is:

 

2. The total current I is the sum of the individual currents through the resistors:

    I total = I1 + I2 + I3 + I4 + ... + In.  This is called Kirchhoff's Current Law or KCL.

 

Since the voltage V is common to all components, we have a third principle by which:

 

3. V = V1 = I1*R1 = V2 = I2*R2 = V3 =I3*R3 = V4 = I4*R4 = ... Vn = In *Rn.

 

There must be a way to combine all the resistors in a parallel circuit such that our fourth principle:

 

4. V = I * Req, where Req is the equivalent resistance of the circuit holds true by Ohm's Law.

 

To obtain the equivalent resistance we use our second principle such that:

 I = I1 + I2 + I3 + I4 + ... + In        and

V/Req = V1/R1 + V2/R2 + V3/R3 + V4/R4 + ... +Vn/Rn      and by principle 1:

V/Req =  V/R1  +  V/R2  +  V/R3  + V/R4  +  ... + V/Rn      and dividing both sides by the voltage V:

Our fifth principle is that the equivalent resistance can be obtained by:

 

5.  1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4 + ... 1/Rn    or:

    Req  = 1/ (1/R1 + 1/R2 + 1/R3 + 1/R4 + ... 1/Rn)

 

 

 

II. Some Basic Examples of Parallel Resistive Circuits

 

1. Two Resistor Parallel Circuit

    R1 = 100 Ohms and R2 = 20 Ohms are the two parallel

   

    resistors in the circuit.  The current will split into I1 and I2

 

↓  both with the direction of the arrow on the left.  The power

 

    supply voltage V= 10 volts is shared by R1 and R2.

 

The equivalent resistance Req can be solved by using 1/Req = 1/R1 + 1/R2  = ( R2 + R1)/R1*R2.

 

Hence Req is the inverse of the above or Req = R1*R2/(R1 + R2), or in general for two resistors we

 

can say that the equivalent resistance Req is the product over the sum of the resistances.

 

For our specific example: Req = 100*20/120 = 16.67 Ohms.

 

Since V1 = V2 = V, we can easily find I1 and R2 using principle 3 such that : I1 = V/R1 = 10/100 =0.1A

 

and I2 = V/R2 = 10/20 = 0.5A  The total current is I = I1 + I2 (principle 2)  = 0.1A + 0.5A = 0.6A. 

 

Note also that by principle 4   I = V/Req = 10/16.67 = 0.6 A which checks with the previous result.

 

 

2. Three Resistor Parallel Circuit

 

 

        Find all the currents I1, I2, I3 and I = Itotal.

 

        Also Find Req.

 

 

 

Here again we have the currents I1, I2 and I3 in the direction pointing towards ground.

 

We can find Req in the same way as in the previous example so that 1/Req = 1/R1 + 1/R2 + 1/R3.

 

So: 1/Req = 1/1 + 1/2 + 1/5 (where the resistor values are in KOhms, but have been omitted for simplicity.)

 

To solve this equation we need to find the common denominator of the fractions which in this case is 10.

 

Then, we will have: 1/Req = (10 + 5 + 2)/10 = 17/10.  Hence, by taking the reciprocal of this number :

 

Req =  10/17 =  0.59 KΩ.

 

By principle 3 again we can solve for the currents through the resistors such that  I1 = 10/1KΩ. = 10mA,

 

I2 = 10/2KΩ. = 5mA, and  I3 = 10/5KΩ = 2mA.  Very important to notice  that dividing volts by KΩ will

 

give us mA for the current !!!  The total current I is  I = I1 + I2 + I3 (by KCL)  = 10 + 5 + 2 = 17mA.

 

 

3. Parallel Circuit where "common denominator" of resistances is too large.

 

 

    Again Find all the currents

 

    and find Req in the circuit.

 

 

 

 

In this particular case finding Req in the usual way would mean to find the common denominator of 11KΩ,

 

3KΩ, 1KΩ and 5KΩ which happens to be 165 !!.  Dealing with such an unusually large number poses a

 

problem. 

 

To go around this problem we can use principle 4  by which Req = V/I .  If  the voltage V is known as is our

 

case here we can find I by KCL such that I = I1 + I2 + I3 + I4; the individual currents can be found by using

 

principle 3  such that Ix = V/Rx , x representing any of the currents I1 through I4.  Applying this to the problem

 

at hand:  I1 = 10/1KΩ = 10mA,  I2 = 10/11KΩ = 0.91mA,  I3 = 10/3KΩ = 3.33mA, I4 = 10/5KΩ= 2mA.

 

Thus I = 10mA + 0.91mA + 3.33mA + 2mA = 16.24mA.  And Req = 10/16.24mA = 0.62KΩ.

 

 

But what about if the voltage V is not given?   In this case we use the Total Current Method whereby we

 

assume a current of 1A going through the highest resistor in this case R2.  Then we can calculate the voltage by

 

using Ohm's Law so that  V =  I2*R2 = 1A*11KΩ = 11KV.  Once we have the voltage we can calculate the

 

remaining currents: I1 = 11KV/1KΩ = 11A,  I3 = 11KV/3KΩ = 3.67A,  I4 = 11KV/5KΩ = 2.2A.

 

Adding the currents using KCL:  I = 11A + 1A + 3.67A + 2.2A = 17.87A. 

 

Finally:  Req = V/I = 11KV/17.87A  = 0.62KΩ  which matches the result obtained before !!!

 

 

 

III. Current Dividers

 

    We have several resistors in parallel and we do not know

 

    the voltage across them.  Can we find out any of the currents

 

    in the circuit if we know any other current?

 

    The answer is Yes and the way we do it is by using the

 

    Current Divider Rule.

 

 

The Current Divider Rule is derived from the fact that the voltages are all the same, hence V = V1 =V2 =V3

 

and so on or in other terms:  I*Req = I1*R1 = I2*R2 = I3*R3 and so on.  Thus for any current Ix we can

 

state that  Ix = I*Req/Rx or that the current through any of the branches is equal to the total current times

 

the ratio of the equivalent resistance to the resistance of that particular branch.  This is the Current Divider Rule.

 

 

Circuit Example:

   

    The total current and resistors for the circuit

 

    on the left are given.  The voltage is unknown

 

    Find the currents I1 through I3.

 

 

The equivalent resistance for this circuit was already previously calculated as Req = 0.59KOhm  for simplicity 

 

we shall round this number to 0.6KΩ.

 

In this way the Current Divider Rule tells us that  I1 = I*Req/R1 = 1mA* 0.6/1 = 0.6mA

 

In the same way,  I2 = I*Req/R2 = 1mA*0.6/2 = 0.3mA  and  I3 = I*Req/R3 = 1mA*0.6/5 = 0.1mA

 

To doublecheck by KCL:  I1 + I2 + I3 = 0.6mA + 0.3mA + 0.1mA = 1mA which checks !!!

 

 

IV. Power in Parallel Circuits

 

                                                

        There is nothing particularly special about the power of resistors in a parallel

 

        circuit.  If the 3 Antennas on the left were in parallel the total power would

 

        be the sum of the individual power dissipated through each antenna, the same

 

way as for a series circuit so that  Ptotal = P1 + P2 + P3  and in the most general case we have that:

 

Ptotal = P1 + P2 + P3 +... + Pn   for a total of n resistances in parallel.  Note that  P = I*V also holds

 

as well as the formulas previously studied of power across resistors.

 

 

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