电阻串联和平行Combinations

电阻串联和平行

电阻可以单独使用串联连接或单独连接连接。一些电阻电路是由系列和平行网络组合制成的，以开发更复杂的电路。这些电路通常称为混合电阻电路。即使这些电路具有组合串联和平行电路，但计算等效电阻的方法也没有变化。单个网络的基本规则，例如“相同的电流流过串联电阻”和“平行电阻跨电阻的电压相同”，适用于混合电路。

一个n example of mixed resistors circuit is shown below

它由混合电阻电路组合中的四个电阻R1，R2，R3和R4组成。电源电压为V，电路中流动的总电流为I。流经电阻器R2和R3的电流为I1，流过电阻R4的电流为I2。

在这里，电阻R2和R3以串联组合为组合。因此，在串联组合中应用电阻规则，R2和R3的等效电阻为

r_一个= R2 + R3

这里R一个is the equivalent resistance of R2 and R3

Now the resistors R2 and R3 can be replaced by a single resistor RA. The resulting circuit is shown below.

Now the resistors RA and R4 are in parallel combination. Hence by applying the rule of resistors in parallel combination the equivalent resistance of RA and R4 is

r_b= R_一个×r4 /（r_一个+ R4）

在这里，RB是RA和R4的等效抗性

现在，我们可以用单个电阻RB替换电阻RA和R4。替换电阻后，所得电路如下所示。

Now the circuit consists of only two resistors. Here also the resistors R1 and RB are in series combination. Hence by applying the rule of resistors in series the total circuit equivalent resistance is given as

r_等式= R1 + R_b

这里R_等式is the total circuit equivalent resistance. Now the resistors R1 and R_b可以用单个电阻R替换_等式。

上面的复杂电路的最终等效电路如下所示。

尽管它们看起来很复杂，但混合电阻电路可以通过遵循串联电阻和电阻的简单规则和并行的电阻规则来简单电路，该电路仅由一个电压源和单个电阻组成。

串联和平行示例中的电阻器

Let us calculate the equivalent resistance for the below circuit which consists of 7 resistors R1 = 4 Ω, R2 = 4 Ω, R3 = 8 Ω, R4 = 10 Ω, R5 = 4Ω, R6 = 2 Ω and R7 = 2Ω.The supply voltage is 5 V.

现在，电阻R6和R7是串联组合的。如果R6和R7IN系列的等效电阻为RA，则

RA = R6 + R7 = 2 + 2 =4Ω

将最终的电路还原为下面所示的电路。

In the above circuit the resistors Ra and R5 are in parallel combination. Hence the equivalent resistance of Ra and R5 is

r_b= (R_一个× R₅) / (R_一个+ R₅) = (4 × 4) / (4 + 4) = 2Ω.

然后显示了简化的电路。

在此电路中，电阻R4和R_b一个re in series combination.

RC = R4 + R_b= 10 + 2 =12Ω。

Now we can replace the resistors R4 and R_bwith resistor Rc as shown below.

In the above circuit again the resistors R2 and R3 are in series combination. If Rd is the equivalent resistance of R2 and R3then

RD = R2 + R3 = 4 + 8 =12Ω。

the equivalent circuit is

这里电阻RC和RD平行组合。令RP成为RC和RD并行的等效电阻。然后

r_p= (R_C× R_d) / (R_C+ R_d）=（12×12） /（12 + 12）=6Ω。

最终的电路是

在这里，电阻R1和RP是串联组合。令r_等式be the equivalent resistance of this combination.

然后

r_等式= R1 + Rp = 4 + 6 = 10 Ω.

这是电路的等效电阻。因此，给定电路最终可以重新绘制为

电路中的电流可以根据欧姆定律计算

i = v / r_等式= 5 / 10 = 0.5 A

resistor Network

Let us calculate the equivalent resistance for a complex resistor circuit.

以下电路由十个电阻器R1到R10组成，这些电阻与系列和平行连接的组合相连。

the values of the resistances mentioned in the circuit are in Ohms (Ω) and the supply voltage is in Volts (V).

Here the resistors R9 and R10 are in series combination. Let R_一个是这种组合的等效抗性。

因此r_一个= R9 + R10 = 3 + 3 = 6 Ω.

the circuit after replacing R9 and R10 with R_一个is

In this circuit, the resistors R8 and R_一个并行组合。然后the equivalent resistance of R8 and R_一个is

r_b=（r8×r_一个) / (R8 + R_一个) = (6 × 6) / (6 + 6) = 3 Ω.

Now replacing R8 and R_一个与r_b, we get the following circuit.

In this circuit, the resistors R7 and R_b一个re in series combination.

r_C= R7 + R_b= 9 + 3 = 12 Ω.

the equivalent circuit after replacing R7 and R_b与r_Cis

显然，电阻R6和RC是平行组合的。如果r_dis the equivalent resistance of this combination, then

r_d= (R6 × Rc) / (R6 + Rc) = (12 × 12) / (12 + 12) = 6 Ω.

R¬D取代R6的电路和Rc

现在电阻R4和RD串联组合。如果RE是R4和RD的等效电阻，则

r_e= R4 + R_d= 6 + 6 =12Ω。

the resulting reduced circuit after replacing R4 and R_d与r_eis

在此电路中，电阻器R5和R_e并行组合。

令r_F成为R5和R的等效电阻_e在平行下。

然后

r_F=（r5×r_e) / (R5 + R_e）=（12×12） /（12 + 12）=6Ω。

the simplified circuit is as shown below.

在这里，电阻R2和R3串联。如果rg等同于该组合，那么

r_G= R2 + R3= 4 + 2 = 6 Ω.

一个fter replacing R2 and R3 with RG, the circuit will be transformed to

电阻RF和RG平行。

令r_t相当于这种组合。

然后r_t= (R_F× R_G) / (R_F+ R_G) = (6 × 6) / (6 + 6) = 3 Ω.

现在电阻R1和RT串联。如果REQIS总电路当量电阻，则REQ = R1 + RT = 3 + 3 =6Ω。

Finally the above complex circuit can be redrawn as follows

电路中的总电流可以使用欧姆定律计算

i = v1 / r_等式= 6 / 6 = 1 A

Hence any complex resistive circuit consisting of number of resistors connected in combination of both series and parallel combinations can be reduced by first identifying the simple parallel resistor branches and series resistor branches. The equivalent resistance of these simple branches is calculated and the branches are replaced with the equivalent resistor. This process reduces the complexity of the circuit. By continuing this process we can replace a complex resistive circuit with a single resistor.

there are some complexresistive circuits通过简单地应用串联电阻组合和平行电阻组合的规则，不能将其简化为简单的电路。诸如T-PAD衰减器和某些复杂的电阻桥网络之类的电路是这种复杂的电阻电路的示例。为了简化这些复杂的电阻电路，将遵循另一种方法。

通过使用Kirchhoff的现行法律和Kirchhoff的电压法，可以降低一些复杂的电阻电路。

to find the currents and voltages in a complex resistive circuit just by using Ohm’s law might not possible. For such type of circuits Kirchhoff’s Circuit Laws will be helpful.

Kirchhoff的电路法律是基于电路中电流和能源保护的概念。有两项基尔乔夫的巡回法则。首先是基尔乔夫（Kirchhoff）的现行法律，该法律涉及Node目前的法律，其次是Kirchhoff的电压法，该法律处理闭路电路中的电压。

Kirchhoff’s Current Law states that “The current entering a node is equal to the current leaving the node because it has no other place to go and no current is lost in the node.”

In simple words, the Kirchhoff’s Current Law states that the sum of currents entering a node is equal to sum of currents leaving the circuit.

Kirchhoff的电压法指出：“闭环中的总电压等于该循环中所有电压降的总和。”

In simple words, the Kirchhoff’s Voltage Law states that the directed algebraic sum of voltages in a closed loop is equal to zero.

借助这两个定律，可以计算任何复杂电路中电流和电压的值。

Still we may have some complex resistive circuits in which it is difficult to identify the equivalent resistance, in such situations we will use Star Delta Transformations of resistors in order to simplify the resistor networks.