The problem of the resistor circuit is very urgent, please master, thank you in advance

Grounding Point: v2-= 0

then v2+ = bat2

V1+ and V2+ are the same point, the voltage is the same V1+ = BaT2, then V1- = V1+ -BaT1 = BaT2 - BaT1 diode D1 on-voltage (silicon, germanium is constant, assuming Vd1 = then V3 = V1- +Vd1 = BaT2 - BaT1 +V0 is the partial voltage value of R3 and R4 in series, V0 = [R4 (R3+R4)] V3 = [R4 (R3+R4)] BaT2 - BaT1 +

This graph scores two scenarios to calculate:

1. D1 cut-off.

When the voltage drop of the current generated by VBAT2 on R2 is greater than or equal to that of BAT1, D1 cuts off. (When considering the forward pressure drop of D1, please calculate it by yourself).

vr2=vbat2*r2 (r2+r3+r4) vbat1, i.e., vbat2 vbat1*(r2+r3+r4) r2.

At this time, vo=vbat2*r4 (r2+r3+r4)2, d1 is on.

When the voltage drop between the current generated by VBAT2 on R2 is less than that of BAT1, D1 is turned on. (When considering the forward pressure drop of D1, please calculate it by yourself).

i.e. vbat2 at this time the voltage across r2 is equal to vbat1.

vo=(vbat2-vbat1)*r4/(r3+r4)

The general idea is this, first of all, let's assume that the diode is removed, and the V3 voltage is V2 (R2+R3+R4)*(R3+R4).

And v1 -=bat2-bat1

Depending on the size of v3 and v1- at this time, there are two scenarios.

When v3>v1-+, according to the simplification of the above formula, we can get:

When bat1>bat2*r2 (r2+r3+r4)+ when the diode is turned on, then the v3 voltage is limited to v1-+, then vo=v3 (r3+r4)*r4=(bat2-bat1+

When bat1< bat2*r2 (r2+r3+r4)+, the diode is not on, then vo=bat2*r4 (r2+r3+r4)

Let me answer this question, because the answer of the previous one is basically wrong.

R13: You can say that the current limiting resistor and the protection resistor can be just without the kind of statement he said, because it is possible that the short circuit can also work normally. The reason why it is used instead of short circuit is to prevent the circuit from burning out the integrated circuit abnormally.

Second, there is no problem with this circuit being set up like this, because he only saw that the VCC was marked on the drawing, if the circuit is a low-voltage circuit such as 5V, and then the single-chip microcomputer power supply is also 5V, isn't this perfect? People's horns are buzzers, and there is no need to think so much about low voltage and low current, and it can be completely saturated and conducted.

The y- transformation can be performed, where r2 r3 r4 is a connection with a resistance value of 1, which can be equivalent to a y connection with a resistance value of 1 3.

So the total resistance is .

First of all, determine which side of R4 has a high voltage, assume that R4 is disconnected, according to the series-parallel relationship of R2, R5, R3, R6, you can judge that the voltage at the right end of R4 is high, so no matter how large the resistance value of R4 is, the current is to the left, let R2, R5, R4, R3, and R6 flow through G2, G5, G4, G3, and G6 respectively, and the voltages of the 4 nodes are U1, U2, U3, and U4, and solve them according to Ohm's law.

The "Prince of Mathematics, Physics and Chemistry" method is the simplest, and "Performing Y- Transformation" needs to look at the textbook "Circuit Science".

The questions and answers of the previous people are simply misleading, and there is no way to get the answers.

Let me answer that:

When the resistance of the pencil lead connected circuit part is 0, then the bulb voltage is 3V, and when the resistance of the pencil lead connected circuit part is 22 ohms, then the bulb voltage is 3x11 (11+22)=1(V), here it is noted that the adjustable is the resistance of the pencil lead.

The result is that the bulb voltage range is 1V-3V.

The meaning of this question should be "the length of the shaded part is half the length of the rectangle, and the width of the shaded part is half the width of the rectangle", assuming this is true, then:

The resistance value is proportional to the length in the direction of the current and inversely proportional to the width.

Suppose that the width of this rectangle is x, i.e. the length is 2x, then:

The resistance value of the two points ab is: x x=1

The resistance value of the two points of cd is: 2x (x 2)=4

Conclusion: The ratio of resistance between AB and CD is 1:4

Calculated according to the resistance definition.

You don't give the length and width of the shaded part, and you can't give you an exact answer. Personally, I believe that the area ratio is the resistance ratio.

rab = ohm, the circuit equivalence diagram is as follows:

<> from both ends of the current source, 4 in parallel 12 = 4 12 = 3;

Then series 3, then series 1, 3 + 3 = 1 = 7;

Then connect 14 in parallel: 7 14 = 14 3.

4 and 12 in parallel to 3;

The right branch of the current source is equivalent to 3 and 1, another 3 and the three are connected in series to 7;

This 7 is connected in parallel with the left 14 of the current source.

Therefore, the right branch of the current source (equivalent resistance 7) current = 18 14 (7 + 14) = 12a

12 The upper current is i=12 4 (12+4)=3A.