Figure 1 shows a parallel circuit with three branches. Each current path is a branch. The current from the power supply is divided and flows through each branch in proportion to the resistance of the branch.

**①Parallel current**Figure 2 shows the branch currents of a parallel circuit, with A, B, and C marked below each branch. The voltage (E) output by the power supply drives the current through each branch, and the total current (I

_{T}) is divided in each branch. The current provided to each branch is Ia, Ib, and Ic. The current of each branch can be calculated using Ohm’s law I=E/R. Note that for each branch, the voltage E from top to bottom is equal. Therefore, if a voltmeter is connected across each branch, the same voltage E will be measured. The current of each branch is I

_{a}=E/R

_{a}, I

_{b}=E/R

_{b}, and I

_{c}=E/R

_{c}. At node A, the total current flows out of current (I

_{a}) and flows through branch A. At node B, the current (I

_{b}) is shunted and flows through branch B. At node C, the remaining current flows through branch C. The total current output by the power supply is shunted according to the ratio of the resistance of each branch, and I

_{T}=I

_{a}+I

_{b}+I

_{c}. This relationship is called Kirchhoff’s current law. Remember that the currents of each branch will accumulate, and the current returning to the power supply is equal to the current flowing from the power supply.

**②Parallel voltage**In a parallel circuit, the voltage drops at both ends of all parallel branches are equal. In other words, the voltage drop on the branch is equal to the voltage applied to the branch. The three circuit connection modes in Figure 3 are functionally equivalent. In Figure 3a, Figure 3b and Figure 3c, the voltage at both ends of branch A-B is equal to the voltage at both ends of branch C-D, and they are all equal to voltage E. If E=12V, the voltage drop of each branch is 12V, and E=VR

_{1}=VR

_{2}=12V.

**③ Ohm’s law in parallel circuit**Apply Ohm’s law to calculate the branch currents in Figure 3a, Figure 3b, and Figure 3c.

Example 1

(1) Calculate each branch current and total current in Figure 3a. R

_{1}=20Ω, R

_{2}=20Ω, E=12V

I

_{1}=E/R

_{1}=12/10A=1.2A

I

_{2}=E/R

_{2}=12/20A=0.6A

I

_{T}=I

_{1}＋I

_{2}=1.2＋0.6A=1.8A

(2) Calculate each branch current and total current in Figure 3b. R

_{1}= 20Ω, R

_{2}=10Ω, E=12V.

I

_{1}=E/R

_{1}=12/20A=0.6A

I

_{2}=E/R

_{2}=12/10A=1.2A

I

_{T}=I

_{1}＋I

_{2}=0.6＋1.2A=1.8A

(3) Calculate the branch current and total current in Figure 3c. R

_{1}=10Ω, R

_{2}=30Ω, E=12V.

I

_{1}=E/R

_{1}=12/10A=1.2A

I

_{2}=E/R

^{2}=12/30A=0.4A

I

_{T}=I

_{1}＋I

_{2}=1.2A＋0.4A=1.6A

Please note that although the three circuits in Figure 3 look different, only the branches are “twisted” into different shapes, and they are actually parallel circuits. In each circuit, the path of current flows from the (-) end of the power supply through nodes B and A and nodes D and C back to the (+) end of the power supply. The total currents of Figure 3a and Figure 3b are equal because the sum of the currents of the branches in each figure is equal. In Figure 3c, the resistance (R_{2}) increases and the total current decreases. This shows that the change of branch resistance will change the size of the total current.

Example 2

In the three branches shown in Figure 4, R_{1}=20Ω, R_{2}=30Ω, R_{3}=60Ω, and V_{3}=120V.

Calculate V_{1}, V_{2}, E, I_{1}, I_{2}, I_{3} and I_{T}

(1) R_{1}, R_{2} and R_{3} are connected in parallel, so the voltage at both ends of R_{1}, R_{2} and R_{3} is equal to E, which is 120V

V_{1}=120V V_{2}=120V V_{3}=120V E=120V

(2) Apply Ohm’s law:

I_{1}=V_{1/}R_{1}=120/20A=6A

I_{2}=V_{2/}R_{2}=120/30A =4A

I_{3}=V_{3}/R_{3}=120/60A=2A

(3) The total current (I) is equal to the sum of the currents of each branch:

I_{T}=I_{1} +I_{2} +I_{3}=(6＋4＋2) A=12A

Suppose the circuit wiring of a certain room is shown in Figure 4 (Example 2), each branch is a power socket, and R_{1}, R_{2} and R_{3} are different electrical loads (such as light bulbs, TV sets or electric fans, etc.). The typical socket voltage is 120V. Each load has a current, and the current flowing through the melt is the sum of the currents in each branch. (Note that the melt is connected in series with all other loads.) Adding or removing a branch will not affect other parallel branches, but will only affect the total current through the melt. If the current exceeds the rating of the melt (or circuit breaker), all branches will be disconnected from the power supply.

Figure 5 shows a simplified room wiring situation. Sockets 1 and 2 are always live, while socket 3 is controlled by switch S_{3}, and bulbs L_{1} and L_{2} are controlled by switches S_{1} and S_{2}, respectively. There are many wiring methods in real applications, but they should all follow electrical codes (including international codes or local codes) to ensure that the wiring meets safety requirements.

Example 3

Figure 6 shows a circuit with four branches. Assume that the fourth branch can be connected to or disconnected from the circuit through terminals A and B. R_{1}=20Ω, R_{2}=30Ω, R_{3}=60Ω, E=120V. When terminals A and B are disconnected (no resistors connected), calculate I_{1}, I_{2}, I_{3} and I_{T}; when terminals A and B are connected with resistor R_{4}=100Ω, calculate I_{1}, I_{2}, I_{3}, I_{4} and I_{T}. .

(1) When R_{4} is disconnected (without resistor):

I_{1}=E/R_{1}=120/20A=6A

I_{2}=E/R_{2}=120/30A=4A

I_{3}=E/R_{4}=120/60A=2A

I_{T}=I_{1}＋I_{2}＋I_{3}=(6＋4＋2)A=12A

(2) When R_{4}=100Ω:

I_{1}=E/R_{1}=120/20A=6A

I_{2}=E/R_{2}=120/30A=4A

I_{3}=E/R_{3}=120/60A=2A

I_{4}=E/R_{4}=120/100A=1.2A

I_{T}=I_{1}＋I_{2}＋I_{3}＋I_{4}=(6＋4＋2＋1.2)A=13.2A

Note that regardless of whether R4 is connected to the circuit, the current through R_{1}, R_{2} and R_{3} will not change, but when R_{4} is connected to the circuit, I_{T} will increase. That is, when the load is connected to the circuit, the total current will increase.

**④Total resistance in parallel circuit**Figure 7a shows a series circuit with a load with a resistance of 10Ω and a current of 12A. When the resistance increases (see Figure 7b), the current will decrease. This result is in full compliance with Ohm’s law, that is, when the voltage does not change, the resistance increases and the current decreases.

When R_{1}=10Ω, E=120V, calculate I_{T}.

I_{T}=E/R_{1}=120/10A =12A

Now, when R_{1}=10Ω, R_{2}=20Ω, E=120V, calculate I_{T}.

I_{T}=E/(R_{1}＋R_{2})=120/(10 +20) A= 120/30A =4A

The result shows that when R_{2} and R_{1} are connected in series, the current I_{T} will decrease.

Now change the resistor R_{2} in Figure 7b to parallel, as shown in Figure 8, and look at the impact on the current. Among them, R_{1}=10Ω, R_{2}=20Ω, E=120V, calculate I_{1}, I_{2} and I_{T}.

I_{1}=E/R_{1}=120/10A=12A

I_{2}=E/R_{2}=120/20A=6A

I_{T}=I_{1}＋I_{2}=12A＋6A=18A

Compare the current in the circuit shown in Figure 8 and Figure 7b. Each circuit uses resistors of the same resistance, and the voltage E remains the same, but in the parallel circuit, the current reaches 18A, while in the series circuit, the total current is only 4A, so the resistance value of the parallel circuit is higher. small. In a parallel circuit, when the branch increases, the total current will increase. This does not seem to comply with Ohm’s law (I=E/R), because the law states that when the resistance increases, the current should decrease. But in fact, when the parallel load increases, the total resistance will decrease, so the current will increase, which is consistent with Ohm’s law.

Imagine a highway with only one exit ramp leading to destinations A, B, C, and D. As shown in Figure 9a, cars must pass through each destination in series, which reduces the total traffic flow. In Figure 9b, each outlet leads to a destination, which is equivalent to a parallel configuration. When the parallel paths are increased, the resistance of the traffic will decrease, so the traffic flow will increase. Therefore, in a parallel circuit, adding branches will reduce resistance and increase current.

**⑤Calculate the total parallel resistance**When calculating the resistance of two parallel resistors, first calculate the product of the two resistances, and then divide by the sum of the resistances (see Figure 10).

R

_{T}= (R

_{1}×R

_{2})/(R

_{1}+R

_{2})

Example 4

Calculate the total resistance of the circuit shown in Figure 11.

R_{T}= (R_{1}×R_{2})/(R_{1} +R_{2})=(220×440)/(220＋210)Ω=96800/660Ω=146.7Ω

This method of dividing the resistance product by the resistance sum is only applicable when two resistors are connected in parallel. In order to calculate the total resistance of any number of resistors in parallel, the reciprocal method shown in Figure 12 can be used. To calculate the total resistance, first calculate the sum of the 1/R term, and then find the reciprocal of this sum. (Note: The reciprocal of a number is equal to 1 divided by this value. For example, the reciprocal of 100 is 1/100, which is equal to 0.01.)

This method is suitable for the calculation of any number of resistors in parallel, and of course it is also suitable for the case of two resistors.

Example 5

Calculate the total resistance of the parallel circuit shown in Figure 13, and draw an equivalent series circuit diagram.

R_{T}=1/(1/2000 +1/3000 +1/800 + 1/4000)Ω

=1/(0.00050 +0.00033 +0.00125 +0.00025)Ω

=1/0.002433Ω=429.18Ω

The equivalent series circuit of the parallel circuit, that is, the voltage E of the parallel circuit is connected in series with the total resistance of the parallel circuit, as shown in Figure 14. The total current of the parallel circuit is equal to the current of its equivalent series circuit.

Example 6

For the circuit shown in Figure 15, calculate R_{T}, E_{T}, V_{1}, V_{2}, V_{3}, I_{1}, I_{2}, and I_{3}.

(1) Total resistance

R_{T}=1/(1/R1 +1/R2 +1/R3)=1/(1/1000 +1/2200 +1/4300)Ω

=1/(0.001 +0.00045 +0.00023) 2=1/0.00168Ω

=595.23Ω

(2) Voltage E

E_{T}=I_{T}×R_{T}=0.08064×595.23V =47.99V

(3) Voltage V_{1}, V_{2} and V

V_{1}=V_{2} =V_{3}=E_{T}=47.99V

(4) Current I_{1}, I_{2} and I_{3}

I_{1}=V_{1}/R_{1} =47.99/1000A =0.0479A

I_{2}=V_{2}/R_{2} =47.99/2200A =0.0218A

I_{3}=V_{3}/R_{3}=47.99/4300A =0.0112A)

Note that I_{1}+I_{2}+I_{3}=80.9mA, which is slightly different from the total current in the figure, which is caused by rounding errors in the calculation process.

In Examples 4, 5, and 6, the total resistance (R_{T}) is smaller than the smallest branch resistance. As a general rule, the total parallel resistance is always smaller than the smallest branch resistance. If the calculated total resistance is relatively large, then you should check whether the calculation process is correct.

**⑥Total power (watts)**The total power (P

_{T}) is the sum of the power of each branch, which can be calculated by one of the following power formulas:

P=V×I

P=V²×R

P=I²×R

P

_{T}=P

_{1}＋P

_{2}＋P

_{3}＋…＋P

_{n}

(P

_{n}is the power of the last branch.)

Example 7

Calculate the power of each branch and the total power of the circuit in Figure 16. E=50V, R_{1}=1kΩ, R_{2}=2.2kΩ.

(1) Branch power

P_{1}=V_{1}²/R_{1}=(50 ×50)/1kQ=2500/1000 W =2.50W

P_{2}=V_{2}²/R_{2}=(50×50)/2.2kQ=2500/2200W=1.14W

(2) Total power

P_{T}=P_{1}+P_{2}=(2.5 +1.14)W=3.64W