People often connect ordinary batteries and photovoltaic cells (solar cells) in series to increase the power supply voltage, and resistors in series can be used to reduce the current in the circuit, or as a voltage divider. The general power supply can be either DC or AC. The circuits described below all use DC power.

**①The current in the series circuit**There is only one current path in a series circuit. As shown in Figure 1, all resistors are connected in series, and when electrons leave one resistor, they enter the other resistor. The electrons must flow through all the resistors before they can return to the positive terminal of the battery. Connect the ammeter to any part of the series circuit, as shown in Figure 2, the measured current values are the same. Therefore, the current at each point in the series circuit is the same.

**②Voltage in series circuit**The reason why current can flow in the circuit is driven by the voltage source. If the current and resistance can be measured, then the voltage value that pushes the current through each resistor can be calculated according to Ohm’s law (V=I×R). This voltage is called a voltage drop, because the voltage source must provide enough voltage to drive current through each resistor, and the voltage “drops” when it crosses the resistor.

In a series circuit, the sum of all voltage drops is equal to the power supply voltage. Figure 3 illustrates this principle, where the current (I) flows through the circuit, assuming that the voltage drop on R

_{1}is 3V, the voltage drop on R

_{2}is 6V, and the voltage drop on R

_{3}The voltage drop is 3V, so the sum of the voltage drops is 12V, which is exactly equal to the power supply voltage. The voltage drop across the resistors R1 and R3 is equal, so the resistance values of the two resistors are also equal.

example 1

In Figure 4, E=12V. Assuming V_{a-b}=4V, V_{b-c}=6V, and the current is 20mA, calculate the resistance value of V_{.c-d }and each resistor.

V_{c－d}=E－(V_{a－b}+V_{b－c})=12V－(4+6)V=2V

R_{1}=V_{a－b}/I=4/20 x 1000Ω=200Ω

R_{2}=V_{b－c}/I=6/20 x 1000Ω=300Ω

R=V._{c－d}/I=2/20 x 1000Ω=100Ω

**③Resistance in series circuit**In a series circuit, the current flowing through each resistor is equal, and the current is limited by the total resistance in the circuit. If the voltage is kept constant, every time the resistance in the series circuit increases, the current will decrease, and when the resistance in the circuit decreases, the current will increase. The total resistance in the series circuit is equal to the sum of the resistances of all resistors.

**④The power in the series circuit**If you know one of voltage and current, current and resistance, or voltage and resistance, you can calculate the electrical power in watts. Each load and resistor in the circuit will have an impact on the power of the circuit, and the total power consumed by the circuit is equal to the sum of the power consumed by all loads and resistors.

**⑤The rules for calculating series circuit parameters**The rules for calculating current, voltage, resistance and power in a series circuit are as follows:

(1) The currents at all points in the circuit are equal, and the total current is

I

_{T}=I

_{1}=I

_{2}=I

_{3}=….=I

_{n}

(2) The total voltage E is equal to the sum of the voltage drops across all resistors

E

_{T}=V

_{1}＋V

_{2}＋V

_{3}＋…＋V

_{n}

(Note: This rule is called “Kirchhoff’s Voltage Law”.)

(3) The total resistance Rr is equal to the sum of the resistance values of all resistors

R

_{T}=R

_{1}+R

_{2}+R

_{3}+…+R

_{n}

(4) The total power Pr of the circuit is equal to the sum of the power consumed by all resistors

P

_{T}=P

_{1}+P

_{2}+P

_{3}+…+P

_{n}

Example 2

There are three resistors in series in the circuit, as shown in Figure 5. The circuit parameters are E=120v, R_{1}=10kΩ, R_{2}=5kΩ, R_{2}=15kΩ. Calculate R_{T}, I_{1}, I_{2}, I_{3}, I_{T}, V_{1}, V_{2}, P_{1}, P_{2}, P_{3} and P_{T}.

(Apply Ohm’s law and the rules for calculating series circuit parameters.)

(1) Total resistance R_{T}=R_{1}+ R_{2}+R_{3}=10kΩ +5kΩ+15kΩ=30kΩ

(2) Total current I_{T}=E/R_{T}=120V/30kΩ =0.004A =4mA

(3) The currents I_{1}, I_{2} and I_{3} are equal throughout the circuit, so the current I1=I2=I3=IT=4mA

1n

(4) Voltage drop V_{1} V_{1}=I_{1}×R_{3}=4mA x 10kΩ=40V

(5) Voltage drop V_{2} V_{2}=I_{2}×R_{2}=4mA×5kΩ=20V

(6) Voltage drop V_{3} V_{3}=I_{3}×R_{3}=4mA×15kΩ=60V

Please note that the total voltage drop is equal to the power supply voltage of 120V

(7) Power P_{1} P_{1}= V_{1}×I_{1}=40V×4mA =0.16W

(8) Power P_{2} P_{2}=V_{2}×I_{2}=20V×4mA =0.08W

(9) Power P_{3} P_{3}=V_{3}×I_{3}=60V×4mA =0.24W

(10) Total rate P_{T}=P_{1}+P_{2} +P_{3}=0.16W+0.08w +0.24W =0.48W

Another way, you can also use the following formula to calculate PT

P_{T}=E_{T}×I_{T}= 120 x4w=0.48W

Example 3 The circuit shown in Figure 6 has two resistors, E=120v, I=1.2A, R_{1}=40Ω, V_{2}=72V. Calculate V_{1}, R_{2} and P_{T}

(1) Because E_{T} is equal to V_{1} + V_{2} (the principle of a series circuit), V_{1} is equal to the total voltage (E_{T}) minus V2:

V_{1}=E_{T}－V_{2} =120V-72V=48V

(2) The application of Ohm’s law has

R_{2}=V_{2}/I_{2}=(72/1.2Ω)=-60Ω

(3) The total power is

P_{T}=E_{T}×I_{T}=120 ×1.2W=144W

**⑥Polarity of pressure drop**The so-called polarity of the voltage drop on the resistor refers to the positive or negative state of one end of the resistor relative to the other end when a current flows through the resistor, as shown in Figure 7. For resistor R

_{1}, the A terminal is negative relative to the B terminal, which is marked with the symbols “-” and “+” respectively. The voltage drop polarity of the resistor R

_{2}is also marked, that is, the C terminal is negative relative to the D terminal. The polarity can be judged by observing the direction of the current. Because the current flows from the negative end to the positive end, the end where the current enters the resistor is marked with a negative sign, and the end where the current flows out of the resistor is marked with a positive sign.

When measuring DC voltage with an analog voltmeter, it is very important to determine the polarity, otherwise the pointer of the meter will deflect in the reverse direction and even cause damage to the meter head. Similarly, when the components are connected to the circuit, if the polarity is wrong, it may also cause damage. For example, a battery has positive and negative poles. When connecting to a load, you must pay attention to the correct polarity to ensure that the circuit can work normally. The battery polarity symbols are marked on the casings of portable devices such as cameras, radios and mobile phones.

When measuring voltage, we always compare one point with another. In Fig. 7, point A is negative relative to point B, and point C is negative relative to point D.

**⑦Measure the voltage to ground**The circuit in Figure 8 uses ground (Gnd) as the reference terminal. When making a measurement, the ground is at zero potential relative to the other points in the circuit. When measuring voltage in a circuit, one of the test leads of the meter is often clamped to the ground, and the other test lead is connected to the measured point in the circuit.

Assuming that in Figure 8, E is 75V, the voltage V_{a-b} on the resistor is 25V, V_{b-c} is 25V, and V_{e-d} is 25V. If you connect a test lead of a voltmeter to Gnd, the voltage from point d to Gnd is 0V. The polarity at both ends of each resistor can be identified according to the direction of the current. The measurement results of the relative ground terminal are as follows: V_{a-Gmd} is +75V, V_{b-Gmd} is +50V, V_{c-Gmd} is +25V, and V_{d-Gmd} is 0V. Please note that in this circuit, the voltage of the opposite ground terminal measured each time is a positive value.

Example 4

This example is shown in Figure 9. The reference ground is located between resistors R_{2} and R_{3}. The power supply E is 75V, assuming that the voltage drop across each resistor is 25V. Please note that the voltages at points a and b are positive with respect to ground. Point c is the ground potential, and the voltage at point d relative to the ground is negative. What are the following measurement voltages: V_{a—Gnd}, V_{b—Gnd}, V_{c—Gnd}, V_{d—Gnd}?

V_{a—Gnd} = +50V

V_{b—Gnd }= +25V

V_{c—Gnd}=0V

V_{d—Gnd}=-25V (point d is negative relative to the ground)

Please note that in this circuit structure, there are both positive and negative voltages relative to the ground.

**⑧The series connection of power supply**Power supplies are often used in series. If the power supplies are added in series, the voltages should also be added in series. In the additive series, the positive pole of the first power supply is connected to the negative pole of the second power supply, that is, connected end to end. Figure 10 shows the addition of two batteries in series. The positive pole (+) of power supply 1 is connected to the negative pole (-) of power supply 2, and the voltage between output terminals A and B is 12V (6V+6V). Terminal A is (-) and terminal B is (+). Power supplies in series are usually connected in an additive series.

When the power supply is subtracted in series, the voltage should be subtracted. In subtractive series, terminals of the same polarity are connected together. Figure 11 shows the case where two batteries are connected in series. The positive pole (+) of power supply 1 is connected to the positive pole (+) of power supply 2, and the voltage between output terminals A and B is 6V (12V-6V).

Terminal A is (+) and terminal B is (-).

Example 5

Two photovoltaic (PV) arrays A and B are connected in series, as shown in Figure 12. The rated voltage of each array is DC6V and the rated current is 2.5A. Please observe the connection polarity of the array and the load. If the load can consume the rated current of all photovoltaic arrays, what are the load current and voltage?

(1) In a series circuit, the current at each point is the same. Therefore, the load current (I_{L}) is I_{L}=I_{A}=I_{B}=2.5A

(2) This example is the additive series mode, so there are

E_{L}=E_{A}+E_{B}=6V +6V=12V