7- Series Circuits
Objectives:
1. Use ohm’s law to find the current and voltages in a
series circuit.
2. Apply Kirchhoff’s voltage law to a series circuit.
Theory:
In a series circuit, the circuit elements with only one path for
current. Whenever we connect resistors in series, the total
resistance increases, the total resistance of a series circuit is
the sum of the individual resistors. Figure 7-2 illustrates a
series circuit with two resistors. The total resistance is:
RT = ∑𝒏𝒊=𝟏 𝑹𝒊
RT = R1 + R2
The current is the same throughout a series circuit so:
I T = I 1 = I2 = I n
The voltage drops across the resistors:
E = V1 + V2 +…….. + Vn
Apply Ohm's Law:
E = I1R1 + I2R2 + InRn
E = Itotal (R1 + R2 +Rn)
E = Itotal Rtotal
Kirchhoff’s voltage law: the algebraic sum of all voltage
rises and drops around any single closed loop in a circuit is
equal to zero.
It is necessary to take into account the polarity of the
voltage. Current from the source creates a voltage drop
across the resistors. The voltage drop across the resistors
will have an opposite polarity to the source voltage as
illustrated in figure 7-2. We may apply Kirchhoff’s voltage
law by using the following rules:
1. Choose an arbitrary starting point. Go either clockwise
or counterclockwise from the starting point.
2. For each voltage source, write down the first sign you
see and the magnitude of the voltage.
3. When you arrive at the starting point, equate the
algebraic sum of the voltages to zero.
Materials needed:
Four Resistors: R1 = 1kΩ, R2 = 2.2kΩ, R3 = 330Ω, R4 =
1.5KΩ.
One dc ammeter, 0-10 mA.
Procedure:
1. Measure each resistor and record the measured value in
table 7-1. Compute the total resistance for a series
connection. Enter the results in table 7-1.
2. Connect the resistors in series as illustrated in figure7-3.
Then measure the total resistance of the series
connection and verify that it agrees with your computed
value. Enter the values in table 7-1.
Component
Listed
Value
Measured
Value
R1
R2
R3
R4
RT =
1kΩ
2.2kΩ
330Ω
1.5kΩ
5.03kΩ
0.988 kΩ
2.178 kΩ
330.12 Ω
1.498 kΩ
4.991 kΩ
Table 7-1
3. Complete the circuit shown in figure 7-4. The ammeter
is connected in series; compute the current in the circuit
by substituting the source voltage and the total
resistance into ohm’s law.
IT = E/RT
Record the computed current in table 7-2.
Computed
Value
2.982 mA
3.13 V
6.886 V
1.032 V
4.695 V
IT
VAB
VBC
VCD
VDE
Measured
value
3.13 mA
3.103 V
6.839 V
1.036 V
4.713 V
Table 7-2
4. In a series circuit, the same current flows through all
components (proof of this). Compute VAB by multiplying
the total current in the circuit by the resistance between
A and B. Record the result as the computed voltage in
table 7-2.
The voltage drops across the resistors:
E = V1 + V2 +…….. + Vn
Apply Ohm's Law:
E = I1R1 + I2R2 + InRn
E = Itotal (R1 + R2 +Rn)
E = Itotal Rtotal
So Itotal = I1 = I2 = In
The voltage between A and B:
VAB = Itotal * R1 = 0.00313A * 1 * 103Ω = 31.3 V
5. Repeat step 4 for the others voltage listed in table 7-2.
6. Measure and record each of the voltages listed in table
7-2.
VBC = Itotal * R2 = 0.00313A * 2.2 *103Ω = 6.886 V
VCD = Itotal * R3 = 0.00313A * 330Ω = 1.032 V
VDE = Itotal * R4 = 0.00313A * 1.5 *103Ω = 4.695 V
7. Using the source voltage (+15 V) and the measured
voltage drops listed in table 7-2, prove that the algebraic
sum of the voltages is zero. The polarities of voltages are
shown in figure 7-4.
∑𝒇𝒐𝒓 𝒄𝒍𝒐𝒔𝒆𝒅 𝒍𝒐𝒐𝒑𝑨→𝑨 𝑽 = 0.0
15 – 3.103 – 6.839 – 1.036 – 4. 713 = 0
8. Repeat step 7 by starting at a different point in the
circuit.
∑𝒇𝒐𝒓 𝒄𝒍𝒐𝒔𝒆𝒅 𝒍𝒐𝒐𝒑 𝑬→𝑬 𝑽 = 0
(4.713 + 1.036 + 6.839 + 3.103)-15 = 0
9. Vopen = 15 V
Conclusion:
 Whenever we connect resistors in series, the total
resistance increases, and it is the maximum.
 The current is the same throughout a series circuit.
 When we connect resistors in series, the total voltage
equal to the sum of voltages in the circuit.
 Total voltage across the closed loop is equal to zero,
taking into account the electrodes.
 In open circuit, the current value equals zero, and the
voltage value equal source voltage.
For further investigation:
I1 = V1/R1 = 3.13 / 1kΩ = 3.13mA
I2 = V2/R2 = 6.886 / 2.2kΩ =3.13mA
I3 = V3/R3 = 1.032 / 0.33 = 3.127mA
• The current is the same throughout a series circuit.
8 The Voltage Divider
Objectives:
1. Apply the voltage divider rule to series resistive
circuits.
2. Design a voltage divider to meet a specific voltage
output.
3. Confirm experimentally the circuit designed in step 2.
4. Determine the range of voltages available when a
variable resistor is used in a voltage divider.
Theory:
A voltage divider consists of two or more resistors
connected in series with a voltage source. Voltage dividers
are used to obtain a smaller voltage from a larger source
voltage. If you have two equal resistors in series, the
voltage across each will be one-half of the source voltage.
The voltage across any resistor in a series circuit can be
found by finding the total resistance, consider the series
circuit illustrated in figure 8-1, to find the voltage across
R2 , the ratio of R2 to RT is multiplied by the source
voltage. Vi = Vs (Ri/RT), that is the voltage divider formula
can be extended to find the voltage in a series circuit
between any number of resistors.
According to this circuit:
R T = R1 + R2
E = V1 + V2
IT = I1 = I2 = E/RT
Voltage divider rule:
V1 = I1R1 = E* R1 / RT = E * R1 / (R1 + R2)
V2 = I2R2 = E* R2 / RT = E* R2 / (R1 + R2)
V1 + V2 = (E * R1/ (R1+R2)) + (E * R2/ (R1+R2)) = E
So Vi = Ri * E/RT, that is the voltage divider formula.
Materials needed:
Four Resistors: R1 = 1kΩ, R2 = 2.2kΩ, R3 = 330Ω, R4 =
1.5KΩ.
One 1.0kΩ , potentiometer.
Procedure:
1. Obtain the resistors listed in table 8-1. Measure each
resistor and record the measured value in table 8-1.
2. Connect the resistors in the series circuits illustrated in
figure 8-4. With the power off, measure the total
resistance of the series connection and verify that it
agrees with your computed value.
3. Apply the voltage divider rule to each resistor, one at a
time to compute the expected voltage across that
resistor. Use the measured values of resistance and a
source voltage of +10 V. Record the computed voltages
VX in table 8-1.
4. Turn on the power and measure the voltage across each
resistor. Record the measured voltage drops in table 8-1
5. Observe the voltages measured in step 4.
Resistor
R1
R2
R3
R4
Total
Listed
Value
1kΩ
2.2kΩ
330Ω
1.5kΩ
5.03KΩ
Measured
Value
0.988kΩ
2.178kΩ
330.12Ω
1.498kΩ
4.991kΩ
VX= VsRx/RT Vx
Measured
1.979 V
1.974 V
4.363 V
4.350 V
0.661 V
0.659 V
3.001 V
2.996 V
10.004 V
9.979 V
V1 = 10 * 0.988/4.991 = 1.979 V
V2 = 10 * 2.178/4.991 = 4.363 V
V3 = 10 * 330.12 * 10-3/4.991 =0.661 V
V4 = 10 * 1.498/4.991 = 3.001 V
6. Using the 330Ω, 680Ω, and 1.0kΩ resistors, design a
voltage divider with +5.0 V output from a source
voltage of +10 V. draw your design in the space
provided below.
7. Use two of the resistors from this experiment to design a
divider with +10 V input and a 7.5 V output. Draw your
design.
8. The circuit shown in figure 8-3 (b) uses a 1.0kΩ
potentiometer and R1 and R2 to limit the range of
voltages. Assume Vs is +10V. Use the voltage divider
formula to compute the minimum and maximum
voltages available from this circuit.
9. Measure the maximum and minimum output voltages.
VMIN = 1.642 V
VMAX= 4.975 V
Conclusion:
 The voltage across any resistor in a series circuit
can be found by finding the total resistance and
source voltage then calculated using the voltage
divider formula.