Elec. Cir. I Lab    EXP. 3: Series-Parallel Circuits  

This experiment contains links to supplemental information.  Be sure you print that information.

OBJECTIVES:

1. To become more familiar with both series and parallel circuits.

2. To learn how to determine "equivalent resistance" for both series and parallel circuits.

3. To learn how to use "voltage division" and "current division."

4. To become more familiar with applications of Ohm's Law and Kirchhoff's Laws.

BACKGROUND & THEORY

The equivalent resistance of resistors in series is expressed as:

Req = R1 + R2 + R3 ... Rn
The equivalent resistance of resistors in parallel is expressed as:
Note: For only two resistors in parallel, the above equation reduces to:
Note also that for resistors of the same value in parallel this reduces to:

Req = R1/2  for two resistors
Req = R1/3  for three resistors
Req = R1/4  for four resistors
etc.

The principle of voltage division can be used for series circuits, and it is
stated as follows: The total voltage across a circuit of resistances in series will divide itself in the circuit in direct proportion the resistances.


FIG. 1

Using voltage division in the circuit shown in Fig. 1:

Also the voltages in Fig. 1 can be determined by using Ohm's Law, if you know I.
V1 = R1I and V2 = R2I
The principle of current division can be used for parallel circuits, and it is stated as follows: The total current in a circuit of resistances in parallel will divide itself in inverse proportion to the resistances.  Using conductance instead of resistance Where G1 = 1/R1 and G2 = 1/R2 the currents divide in direct proportion to the conductances


FIG. 2

Using current division in the circuit shown in Fig. 2:

or I1 = G1/(G1 + G2) and I2 = G2/(G1 + G2)

Also the currents in Fig. 2 can be determined by using Ohm's Law.


EQUIPMENT AND PARTS LIST:

Digital Multimeter (DMM) }
Resistors:  2 - 1.2 kW
                  1 - 2.7 kW
                  2 - 3.3 kW
                  1 - 4.7 kW
                  1 - 5.6 kW
Adjustable D.C. Power Supply
Circuit Bread Board 

PROCEDURE:

1. Using the adjustable D.C. power supply and circuit bread board, connect the resistors into a circuit conforming to Circuit #1 below. (Be sure to use the color code card to determine that you are using the correct resistors in each position).  Make sure you record the actual value of each resistor used along with the position in which it was used.


CIRCUIT 1:

R1 = 1.2 kW,   R2 = 3.3kW,   R3 = 3.3 kW,
R4 = 2.7 kW,  R5 = 5.6 kW,  R6 = 4.7 kW

2. Measure and record all the currents and voltages in Circuit #1 setting Vs close to 19 V DC.  Measure and record I4 , I5 , & I6 twice.  Once using the 4mA(4000ľA) range and once using the 40mA range.

3. Dismantle Circuit #1 and connect Circuit #2 as shown below.


CIRCUIT 2:

R1 = 1.2 kW,  R2 = 5.6 kW,  R3 = 3.3 kW,
R4 = 4.7 kW,  R5 = 2.7 kW,  R6 = 1.2 kW 

4. Measure all the currents and voltages in Circuit #2 setting Vs close to 20 V D.C.

CALCULATIONS AND COMPARISONS:

1. Using methods presented in BACKGROUND AND THEORY plus the attached reference example, calculate the theoretical values of currents and voltages using the measured values of resistances and Vs for both Circuit #1 and Circuit #2.

2. Make a table to compare measured values with theoretical values and include the % difference for each voltage and current.

3. Apply KVL to each loop and KCL to each node. How closely do the voltages and currents add up to the values predicted?  Which current range gave the most accurate values for I4 , I5 , & I6 in circuit 1?

4. Were Kirchoff's laws verified to within the accuracy of the meter used?  Show how you can demonstrate the overall accuracy of the experiment.

CONCLUSIONS (Based on what you've learned).

 

Reference Example

Example of solving for circuit variables by reducing the circuit to a series of simpler circuits by using series and parallel equivalents.


The first step is to combine R5 and R6 into the series equivalent R56 = R5 + R6 = 5 kW.

V5 and V6 are no longer in the circuit, but V4 and the other voltages are unchanged.  Next we combine R4 and R56 to give the parallel Equivalent R456 = R4*R56/(R4 + R56) = 2.5 kW.

The remaining voltages are still unchanged.  Next we combine the two series resistors R3 and R456 to get R3456=R3+R456=4 kW.

Now V3 and V4 are no longer in the circuit, but V1 and V2 are still unchanged.  Finally we combine R2 in parallel with R3456 giving R23456 = R2*R3456/(R2 + R3456) = 2 kW.

V2 and V1 are still unchanged and we can find both of them by voltage division.  V1 = Vs*R1/(R1 + R23456) and V2=Vs*R23456/(R1+R23456).  This gives V1=3.333V and V2=6.667V.  At this point we can calculate the current through R1.  I1 = Is = I23456 =V1/R1 = V2/R23456 = Vs/(R1 + R23456) = 3.333mA.  or we can calculate I1=Is=I23456= Vs/(R1+R23456) and then calculate V1=I1*R1 and V2=I1*R23456 or V2=Vs-V1.

We now move back to the previous equivalent circuit where we can calculate the current through R2. I2 = V2/R2=1.667mA from Ohm's law or I2=I1*R3456/(R2+R3456) using current division.   The current through R3456 is I3=V2/R3456=1.667mA by Ohm's Law or  I3=I1*R2/(R2+R3456) by current division.

Moving back to the next previous equivalent, We can calculate V3 and V4.  V3=V2*R3/(R3+R456)=2.500V and V4=V2*R456/(R3+R456)=4.1667V, or V3=I3*R3 and V4=I3*R456.

Now moving back again to the next previous equivalent circuit, we  can calculate the current through R4  and the current through R5 and R6.  I4=V4/R4=0.8333mA=I3*R56/(R4+R56).  I5=I3*R4/(R4+R56)=V4/R56

Moving back to the original circuit we can now calculate V5 and V6.  V5=I5*R5 and V6=I5*R6, or V5=V4*R5/(R5+R6)=1.6667V and V6=V4*R6/(R5+R6)=2.500V.

This gives us all the currents and voltages in the circuit without having to solve a set of simultaneous equations.

Last updated 01/30/2008