# Parallel DC Circuits

Next we move onto parallel DC circuits. A parallel DC circuit is one in which the voltage across all the components is the same.  A simple parallel circuit is shown below:

We would like to be able to determine the total current drawn by the circuit, as well as the current in each branch. It would be good if we could extend our equivalent circuit concept, which we discussed in the last section, to parallel circuits, so that we have a way to simplify a complex parallel circuit containing many branches.

We will find the total current in the parallel circuit shown above, ITOTAL, by applying Ohm's Law to each branch. We will use the form of Ohm's Law that allows us to determine the current, that is I = V/R:

I1 = V/R1        current through the branch containing R1

I2 = V/R2        current through the branch containing R2

I3 = V/R3        current through the branch containing R3

These currents are shown in the diagram below.

How do we get from the individual branch currents to the total current? First, look at node N2. The current flowing into the node has to be equal to the current flowing out of it. If that were not true, it would mean that either more electrons were flowing out of the node than were flowing in, which would require that electrons be created out of nowhere or that less current was flowing out than in, and electrons were disappearing into thin air. Neither of these events is possible, and all of this can be stated more concisely as follows: the algebraic sum of the currents entering and leaving a node is zero. This is known as as Kirchoff's Current Law( KCL), and was named in honor of the German physicist Gustav Kirchoff, who discovered it.

Referring to our diagram above:

Current flowing into Node 2 - I2 - I3  = 0    We use a negative sign to indicate that currents are flowing out of the node.

We can simplify this to get:

Current flowing into Node 2 =  I2 + I3

We can apply KCL to Node 1:

Current flowing into Node 1 = I TOTAL - I1 - ( I2 - I3) = 0

Simplification yields:

I TOTAL = I1 +  I2 + I3

We can use our earlier results for I1, I2, and I3 to get the following equation:

I TOTAL = (V/R1) + (V/R2) + (V/R3)

We can simplify this a bit further:

I TOTAL = V[(1/R1) + (1/R2) + (1/R3)]

At this point, one might think that we are finished, but we are not. Let us recall that the conductance of a resistor is the reciprocal of its resistance, that is:

G = 1/R

Using this idea, we can rewrite our equation for the total current in the parallel circuit in a simpler form:

I TOTAL = V(C)

The sum of the individual conductances in the circuit is just the total conductance, GTOTAL. We can simplify our equation even further, to get:

I TOTAL = VGTOTAL

The total current is merely the applied voltage multiplied by the total conductance.

Consider the two circuits below. The one on the right is the equivalent circuit of the one on the left. Remember from our earlier work that the equivalent circuit has the same current flowing for the same applied voltage, but contains only a single resistor.

It would be nice to be able to express Req in terms of R1, R2 and R3, and believe it or not, we know how to do it. We just learned that:

I TOTAL = VGTOTAL

and from this we deduce that the conductance of the equivalent circuit is:

Geq = GTOTAL.

But:

GTOTAL = G1 +G2 +G3 = (1/R1) + (1/R2) + (1/R3)

Also:

GTOTAL = Geq = 1/Req = (1/R1) + (1/R2) + (1/R3)

Combining these results gives:

1/Req = (1/R1) + (1/R2) + (1/R3)

This can be extended (the proof is left to the reader) to a parallel circuit with any number of branches, in which case the formula becomes:

1/Req = (1/R1) + (1/R2) + (1/R3) + ......(1/RN)

where N is the number of parallel branches. This tells us that the reciprocal of the equivalent resistance of a parallel circuit is the sum of the reciprocals of the resistance of each branch.

When a parallel circuit has only two branches, the equivalent resistance of the circuit is given by:

Req = (R1*R2)/(R1 + R2)

Now we will work out some examples.

Example 1:

What are the branch currents and total current in the circuit shown below?

What is the equivalent resistance of the 3 resistors in parallel?

Solution:

To find the branch currents, we apply Ohm's Law:

I1 = V/R1 = 10V/1KΩ = 10V/1000Ω = 0.01 A = 10mA

I2 = V/R2 = 10V/2KΩ = 10V/2000Ω = 0.005 A = 5 mA

I3 = V/R3 = 10V/3KΩ = 10V/3000Ω = 0.0033 A = 3.3 mA

To find the total current, we apply KCL  (add the currents at the node joining the + side of the voltage source, R1, R2 , and R3):

ITOTAL + I1 + I2 + I3 = 0  or ITOTAL = -( I1 + I2 + I3) = -(-10 - 5 - 3.3 mA) = 18.3 mA

Each branch current carries a negative sign, because they are flowing out of the node. The total current has a positive sign, because it flows into the node.

We can find the equivalent resistance two different ways, both of which should give the same answer:

Method 1 - use the formula for equivalent resistance:

1/Req = (1/R1) + (1/R2) + (1/R3) = (1/1K) + (1/2K) + (1/3K) = (1/1000) + (1/2000) + (1/3000)

1/Req =  .001+.0005 + .00033 = 0.00183

Req =  1/0.00183 = 545 Ω

Method 2 - Use the total current that was determined earlier and Ohm's Law:

Req = V/ITOTAL   = 10V/18.3mA = 10/(0.0183) = 545 Ω

Note that the equivalent resistance is less than any of the branch resistances. This is always true and it is a good check on your work. If your equivalent resistance is not smaller than any branch resistance, you have made an error and should recompute the value of Req .

Example 2:

What is the unknown resistance in the circuit below, if the equivalent resistance of the circuit is 278 ohms?

Solution:

We can use either formula for the equivalent resistance of a parallel circuit. We will use the formula containing the reciprocals, because it will be easier to use:

1/Req = (1/R1) + (1/R2)

We know  Req = 278 ohms and one of the resistances is 470 ohms. Thus:

( 1/278) = (1/470) + 1/R or 0.00360 = 0.00213 + 1/R

We can simplify this and solve for 1/R:

1/R = 0.00360 - 0.00213 = 0.001470

We take the reciprocal of both sides to get R:

R = 1/0.001470 = 680 Ω

Now we are ready to look at circuits that contain parallel and series branches

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