There is a question about electronic circuits that I would like to ask a master

Is there a reference direction for the voltage source below?

Since the demand is voltage, the junction voltage method can be used directly.

By the junction voltage method, self-conducting-interconductivity = current in-current out.

Then: Eq. 1: (1 12 + 1 8) un1 - 1 12 + 1 8) un2 = 20 12

Eq. 2: (1 12 + 1 8 + 1 2) un2 - 1 12 + 1 8) un1 - 1 2 un3 = 1-20 12

Eq. 3: (1 2 + 1 2) un3 - 1 2 un2 = -1-2 The solution is: un1 = 6

un2=-2

un3=-4

Therefore, uoc = un1 - 06v answer.

The formula for current-sourced voltage source: e=is ro, and the formula for voltage source-to-current source is=e ro

On the left, the 20V voltage source is first converted into a current source, then.

Current source is = 20 12A

The 12 ohm resistor is connected in parallel with the 8 ohm resistor, which is equal to the ohm.

It is then converted into a voltage source, e1=is *, in an upward direction, in series with the Euclonian resistor.

The following is unchanged. The direction is the same as e1.

On the right, the current source is converted into a voltage source: e2 = 2 * 2 = 4 V, in the direction upward, opposite to the direction of e1, in series with a 2 ohm resistor.

Therefore, the whole circuit is connected by three voltage sources in series, and the total voltage is equal to 6V, that is, the voltage at both ends of the UOC is 6V.

3 resistors are connected in series, and the total resistance is ohm.

It is expressed by the phasor method.

ui = 10 e^(4t+ pi/9)

Circuit in the original:

Resistance: 60 ohms.

Sense: j w * l = 20j ohms.

Capacitive reactance: 1 jwc = 1 (j * 4* = 250j ohms.

So inductor capacitor shunt impedance = -250J * 20J (20J-250J) = 5000J 230 = -500 23 J ohms 22J ohms.

u0＝re［ui ＊（22j）／（60－22j）］

re［10e^(4t+ pi/9) ＊22j）（60＋22j）／（60＾2＋22＾2）］

re［10e^(4t+ pi/9) ＊22j）（60＋22j）／（60＾2＋22＾2）］

re［10e^(4t+ pi/9)（0．12－0．32j）］

10cos(4t+ pi/9) ＊0．12 ＋10sin(4t+ pi/9)0．32

Use additional incentives to seek:

A: External excitation i=i+i*3 6=

uab=3*i+6i=9i

ro=u i=6 ohms.

Figure B: External excitation I

uab=3*i+5*(

ro=u i=5 ohms.

Before t=0, uc(0)=2v;

t= after uc( )=us;

Time constant = c*r;

Substituting the full response formula yields uc(t);

Resistance voltage: ur(t), then there is ur(t) +uc(t) = us;

Resistance current: i(t) = ur(t) r;

1. According to the PN junction current equation:

The relationship between voltage u and current at both ends of the diode is:i=is*((e (qu kt))-1); where is is the reverse saturation current, and each diode has a known value. Kt Q=26mV at room temperature; According to what you said, i is also a known quantity.

At that time, id is the smallest, so the above equation becomes 1mA=is*((E (U 26mV))-1), so that you can find u.

This is a full-wave bridge rectifier circuit, and the basic principle is as follows.