Whether the superposition principle is suitable for nonlinear circuits, and explain the reasons

Whether the superposition principle is suitable for nonlinear circuits, and explain the reasons

The superposition theorem is not applicable to nonlinear circuits. Reason: in linear circuits, r = u / I, so different voltages have different currents when different power sources act. If the voltage (current) when each power source acts alone is added to each component, the current (voltage) of each component can be obtained. After superposition, it is equal to the current (voltage) when it acts together

Calculation of current by superposition principle

When the voltage source acts alone, the current source is open, so I1 = 9V / (4 Ω + 8 Ω) = 3 / 4A
When the current source acts alone, the voltage source is short circuited, so the voltage on 8 Ω = 3A * (4 Ω / / 8 Ω) = 3A * (8 / 3) Ω = 8V, I2 = 8V / 8 Ω = 1a
I = I1 + I2 = 3 / 4A + 1A = 7 / 4A = 1.75A

The superposition theorem is used to calculate the current of each branch and the voltage at both ends of each component, and the power balance relationship is explained

1. When the current source acts alone, the current flowing through 1 Ω, 2 Ω, 4 Ω and 5 Ω is i11, I21, I41 and i51 respectively (the positive direction is set from left to right or from top to bottom). At this time, the 10V voltage source does not work in the circuit, and the two ends of the voltage source are short circuited. Obviously, the 5 Ω resistance is also short circuited, so i51 = 0
At this time, the circuit becomes 1 Ω, 4 Ω resistor in parallel, and then connected to the current source in series with 2 Ω, so I21 = 10A
The 10A current source is shunt by 1 Ω and 4 Ω, so i11 = 4 × 10 / (1 + 4) = 8A; I41 = 1 × 10 / (1 + 4) = 2A
2. When the voltage source acts alone, the current source is open circuit. Set the current flowing through 1 Ω, 2 Ω, 4 Ω and 5 Ω as i12, I22, i42 and i52 respectively (the positive direction is set from left to right or from top to bottom). At this time, the 2 Ω resistor is disconnected from the circuit, so I22 = 0A. 5 Ω resistor is directly connected in parallel at both ends of the 10V voltage source, so i52 = 10 / 5 = 2A
At this time, the circuit changes to 1 Ω and 4 Ω resistors connected in series at both ends of the 10V voltage source, i12 = - 10 / (1 + 4) = - 2A; i42 = 10 / (1 + 4) = 2A
Therefore, when the two power sources work together, the current and voltage of each resistor are:
I1 = i11 + i12 = 8-2 = 6A. U1 = 6V. P1 = 36W, power consumption
I2 = I21 + I22 = 10 + 0 = 10A. U2 = 20V. P2 = 200W, power consumption
I4 = I41 + i42 = 2 + 2 = 4A. U4 = 16V. P4 = 64W, power consumption
I5 = i51 + i52 = 0 + 2 = 2A. U5 = 10V. P5 = 20W, power consumption
For the voltage source, 5 Ω current is 2a from top to bottom, and 1 Ω resistance current is 6A from top to bottom. According to KCl, the voltage source flows in the direction of 6-2 = 4A from top to bottom, so the absorbed power of the voltage source is 4 × 10 = 40W
The voltage at both ends of 4 Ω resistor is 16V, and the voltage at both ends of 2 Ω resistor is 20V, so the voltage at both ends of current source is 16 + 20 = 36V, the direction is from top to bottom, and the released power of current source is 36 × 10 = 360W
The total absorbed and consumed power is 36 + 200 + 64 + 20 + 40 = 360W, and the released power of current source is 360W with energy balance
Note: superposition theorem can be used for current and voltage calculation, but not for power calculation