The question of how to analyze the capacitive circuit and some simple circuits in high school

1. A capacitor is equivalent to a voltmeter. So you can see who you connect with in parallel.

2. According to the voltage difference, judge the direction of the current, 3. The formula of the capacitor is two, learn to use the formula to judge whether the capacitor is charged or discharged, and how the electric field changes. Inductance is an obstacle, not a hindrance. Resisting AC to DC.

There is a mantra to remember whether the ammeter is connected inside or outside, the big inside is big, and the outside is small, learn to understand the process, try not to memorize, and memorize it under the premise of understanding.

It is also a good way to judge the current and voltage, and it is also a good way to understand it as much as possible. In addition, there are some ** that recommend you to learn physics, and it is good to have a simple study network, which is free. In addition, you can find a tutor to teach you to straighten out your thinking and open up your circuit.

That's a big question.

Simplification of electrical circuits.

Principle: 1. The branch without current can be removed.

2. Equipotential points can be merged.

3. The ideal wire can be of any length.

4. The ideal voltmeter is open and the ideal ammeter is short-circuited.

5. When the capacitor is charged, the circuit is broken, which is regarded as parallel connection, and the voltage is equal.

Calculations about capacitors in electrical circuits.

1) The voltage of the capacitor is equal to that of the electrical appliance connected in parallel.

3) When charging and discharging, the current direction on the two leads of the capacitor is always the same, so the current direction should be judged according to the change of the charge of the positive plate.

1) If the electrification of the plates before and after the change is the same, then the amount of charge passing through each lead is equal to the difference between the charge of the capacitor in the beginning and end states; If the electrification of the plates before and after the change changes, the amount of charge passing through each lead is equal to the sum of the charges of the capacitor in the beginning and end states.

Analysis and calculation of capacitive circuits:

Two conductors close to each other, sandwiched between a layer of non-conductive insulating dielectric, constitute a capacitor. When a voltage is added between the two plates of a capacitor, the capacitor stores an electric charge. The capacitance of a capacitor is numerically equal to the ratio of the amount of charge on one conductive plate to the hail voltage between the two plates.

The basic unit of capacitance of a capacitor is the sock farad (f).

In circuit diagrams, the letter C is usually used to denote capacitive components. Capacitors play an important role in tuning, bypassing, coupling, filtering, and other circuits. It is used in the tuning circuit of transistor radios, and it is also used in the coupling circuit and bypass circuit of color television.

With the rapid development of electronic information technology, the upgrading speed of digital electronic products is getting faster and faster, and the production and sales of consumer electronic products based on Yuanhao pants flat panel TVs (LCD and PDP), notebook computers, digital cameras and other products continue to grow, driving the growth of the capacitor industry. A capacitor is a component that stores electricity and electrical energy (electric potential energy).

A conductivity system in which one conductor is surrounded by another conductor, or in which all the electric field lines emitted by one conductor terminate in another conductor, is called a capacitor. Capacitors are not the same as capacitors. Capacitance is a fundamental physical quantity, symbol c, and the unit is f (farad).

The general formula is c=q u, and the special formula for parallel plate capacitors: the electric field strength between the plates is e=u d.

At the moment of closing and closing, the capacitor will be charged, and there will be a short current passing through, the current can either flow from S to the negative electrode through R3CR2, or from R1 to the negative electrode through R2, or SR4 and R1CR3R4, each branch may have current, which involves the appearance of a similar bridge or a star circuit, and it cannot be generalized that this is a series that is parallel.

When S is closed, the voltage at both ends of R1 is equal to the voltage at both ends of the capacitor, and not equal to the voltage at both ends of R2, so when S is closed, R1 is connected in parallel with the capacitor.

(1) If the upper plate of the capacitor is negatively charged, the voltage divided by R4 must be greater than the voltage divided by R2, according to the topic, the voltage at both ends of the capacitor should be 1 volt, so R4 should be adjusted to 20 ohms. If the upper plate is positively charged, the resistance r4 should be adjusted to 4 ohms.

2) 4 to the minus 6 power of 10.

Look at the voltage change of the two segments of the capacitor, and become smaller as a power discharge, and become larger for charging.

Note the question stem "The resistance of the fixed value resistor is r, the number of turns of the coil placed vertically is n, the resistance of the winding coil is also r, and the resistance of other wires is negligible"., the resistance of the winding coil is the internal resistance of the electromotive force e of electromagnetic induction. The equivalent circuit diagram of the circuit is:

The resulting electromotive force e, half of which is distributed to the resistance r of the winding coil, and the other half is divided to a given value of resistance r

Voltage at both ends of the fixed value resistor r: U=mgd q

So e=2u=2mgd q;

The rate of change in magnetic flux is 2mgd nq

Let the current in the circuit be i and the voltage = i r (fixed resistance). Electromotive force = i (r (fixed resistance) + r (coil resistance)).

Not equal, voltage and current are not equal.

Question 17 is the fourth, which is D.

When stable, it can be when R3 is a short wire, so C2 is the power supply voltage and C1 is the voltage of R2.

After C1 and C2 are disconnected, they release their charges, and all charges go R3, so the total amount is Q1+Q2.