电压降公式
电压是使电荷移动的原因,以伏特 (v) 为单位。因此,电压降描述了当电流流过不产生电压的电路部件时,电压源提供的能量如何减少。由于浪费了电源,因此不希望出现跨导体、连接器和电源内阻的电压降。
简而言之,电压降是整个或部分电路发生的电压损失量,因为阻抗称为电压降。
电压降公式
电压降公式可以表示为,
V = I×Z
Where,
- I = the Current which is measured in Ampere,
- Z = Impedance which is measured in ohms.
电压降也以伏特为单位。
示例问题
问题 1:5A 的电流流过电阻为 20 Ω 的闭合电路。计算电路上的电压降。
回答:
Given: Here, we have
- Current (I) = 5A,
- Impedance (Z) = 20 Ω
Thus according to the Voltage Drop formula,
V = I×Z
Substituting the values of Current (I) and Impedance (Z) in the Voltage drop formula,
V = 5 × 20
V = 100 v
Thus voltage drop is 100v
问题2:在闭合电路中,有一个50Ω的电阻和2A的电流。求电流加倍时的电压降。
回答:
Given: Here, we have
I = 2 A
According to the condition, Current is doubled
- I = 4 A
- Z = 50 Ω
Thus according to the Voltage drop formula
V = I × Z
V = 4 × 50
V = 200 v
Thus voltage drop is 200v
问题 3:在闭合电路中,发现电压降为 100 V,电阻为 20 Ω。找出在那个时刻流过电路的电流。
回答:
Given: Here, we have
- Voltage drop = 100 v
- Z = 20 Ω
Thus rearranging the formula for Voltage drop,
I = V / Z
Substituting the value of Voltage drop and Impedance,
I = 100 v / 20 Ω
I = 5 A
Thus the current flowing from the circuit is 5 Ampere.
问题 4:在闭合电路中,电压降为 150 v,电流为 7 A。求阻抗。
回答:
Given: Here, we have
- Voltage drop = 150 v
- I = 7 A
Thus rearranging the formula for Voltage drop,
Z = V / I
Substituting the value of Voltage drop and current
Z = 150 / 7
Z = 21.43 Ω
Thus the impudence offered was 21.43 Ω
问题 5:在一个闭合电路中,两个电阻 R1 和 R2 串联,阻值分别为 15 Ω 和 5 Ω,电流为 6 A。求电路两端的电压降。
回答:
Given: Here, we have
The total resistance in the circuit is R1 + R2 = (15 + 5) Ω = 20Ω
- I = 6 A
- Z = 20 Ω
Thus according to the Voltage drop formula,
V = I × Z
V = 6 × 20
V = 120 v
Thus the voltage drop is 120 volts.
问题 6:在闭合电路中,两个电阻 R1 和 R2 串联,电阻值分别为 10 Ω 和 2 Ω,压降为 80 v。求电路中的电流。
回答:
Given: Here, we have
The total resistance in the circuit is R1 + R2 = (10 +2) Ω = 12Ω
- V = 80 v
- Z = 12 Ω
Thus rearranging the formula for Voltage drop,
I = V / Z
Substituting the value of Voltage drop and Impedance,
I = 80 / 12
I = 6.67 A
Thus the current flowing from the circuit is 6.6667 Ampere.
问题 7:在一个闭合电路中,两个电阻 R1 和 R2 并联,电阻值分别为 8 Ω 和 5 Ω,电流为 4A。求电路两端的电压降。
回答:
Given: Here we have
The total resistance in the circuit is (1/R1 +1/R2) = (1/8 + 1/5) Ω = 0.325 Ω
- I = 4 A
- Z = 0.325 Ω
Thus according to the Voltage drop formula,
V = I × Z
V = 4 × 0.325
V = 1.3 v
Thus the voltage drop is 120 volts.