Circuit calculation problems related to circuits, circuit calculation problems to solve

I don't know what the current knowledge background of the landlord is. I conclude that there are generally four solutions to such problems. The first is the equation listed by the landlord (but the second equation of the landlord is problematic, this will be explained later), which is to use kcl, kvl column to write differential equations, and solve such a problem by solving differential equations.

This solution method is more cumbersome, the complexity of the equation is more related to the selection of variables, and the higher-order differential equations are not easy to solve, so this method is generally not used; The second is to use the three-element method to solve, which is not mentioned here; The third is to perform a complex frequency domain transformation (Fourier transform) and solve it by phasors. The last one, which is more general, is the Laplace transform method. In fact, this method also has two applications, one is to write the equation according to the first column and solve the equation with the Laplace change, and the second is to directly use the Laplace transform to process the circuit and then write the equation. The first two can be regarded as solutions in the time domain, and the last two are generally regarded as solutions in the frequency domain.

Finally, the landlord's problem, the landlord's first equation is not a problem, the second equation has a problem with the resistance part, and the current should be i2-i1. As for solving the equation, the landlord can simply eliminate the element integral.

If you want to understand such similar problems in depth, you can refer to the relevant textbooks of the University School of Telecommunications, such as circuit theory, integral transformation, signals and systems, etc. If you want to know more, you can contact me.

Let me tell you what I think.

Let's think of the power supply and the first inductor as part of the power supply, and let the output voltage be u, so there is:

The total current in the circuit i=u r+u(1-exp(t t)) r=(2-exp(t t))*u r, where t=l r is the time constant.

Next, consider the change of u, and then consider the two resistors and the inductance in the middle as a whole, and their equivalent resistance rx is:

rx=r (2-exp(t t)), i.e., rx tends to r 2 over time.

So by calculating the zero-state response of the L and Rx series circuits, we can directly get an equation where the current in the circuit is:

i=(1-exp(t tx))*e rx, where rx is the result of the above calculation, tx=l rx, so ......It's still a dazzling formula!

System of linear differential equations?

1) The operator method is the easiest way to solve this system of equations, that is, all the equations are transformed into Laplace, so that it becomes an algebraic equation, and then the elephant functions of i1 and i2 are obtained, and finally the inverse transformation is done.

2) There is a classmate above who is actually the elimination method, which can also be used, but it is a bit troublesome to replace it.

Instantaneous value of current

u/r=(311/484)sin(314t+π/6)asin(314t+π/6)a

Method 1: 1 month of electricity = 311 Wh = 12 kWh.

Method 2: The power of the bulb p=(311 2) 484100 watts, and the electricity consumption in 1 month = 4 30 = 120 hours.

1 month Kaiyou internal power belt forest = 100 120 = 12 kWh, that is, 12 degrees.

A1-A3 is the following circuit, and A4 is the in-phase proportional addition circuit grinding core book. Using the principle of superposition, the output of each blind macro input is obtained separately (the figure is the output of the UI1 input), and the roll can be obtained according to the symmetry characteristic

uo=<>

1) Transistor input resistance: rbe = RBB+26(MV) ib;

T1 and T3 are regarded as composite transistors;

rbe1 = rbb+26(mv)/ ib1，rbe2 = rbb+26(mv)/ ib2；

ib2 = ib1（1+β1）,；

Then the total input resistance of the transistor: rbe = rbe1+rbe2;

Total current amplification: 1*2;Total UBE = UBE1+UBE2 = ;

The rest of the calculations are treated as a triode, and then the formula is applied;

This is an RL series circuit, the voltage on each element has a circle of this effect to dress up the relationship of the value of the hall can be represented by a voltage triangle, as follows and two formulas:

From Ohm's law, i=(5-6) 4=, so choose a for this question.

Because of the current i flowing through us2'S2 = Is3 - Is2 = 3A, so, the current flowing through us3.

i's3 = i's2 - is1 = 3a - 3a = 0a

The power of us2 ps2 = -i's2 * us2 = -6w

The power of US3 PS3 = I's3 * us3 = 0w

At this time, the voltage U at both ends of the constant current source IS3's3 = us3 - us2 = 3v

So, the power of the constant current source IS3 = -u's3 * is3 = -15w

Voltage U at both ends of the constant current source IS1's1 = us1 - us3 = 5v

The power of the constant current source is1 = -u's1 * is1 = -15w

The current i flowing into the constant voltage source US1's1 = is1 + is2 = 5a

So, the power of the constant voltage source us1 = us1 * i's1 = 50w

Constant voltage source IS2 voltage at both ends of the voltage U's2 = us1 - u's3 = 7v

Then the power of the constant voltage source is2 = -is2 * u's2 = -14w