How to find the voltage, how to find the power supply voltage

Such a simple topic?? It's really not easy.

u=i*r bar is the easiest.

There are many others, depending on where the voltage is used.

Summary. Hello, according to your related questions, I found for you, the voltage calculation formula: U=W Q. Voltage is a physical quantity that measures the difference in energy of a unit charge in an electrostatic field due to different potentials.

How to do it. Hello Feng Ying shouted, according to your related questions, I Song imitated for you to find out, the voltage calculation formula: U=W Q. Voltage is a physical quantity that measures the difference in energy between a unit of electric silver charge in an electrostatic field due to different electric potentials.

Related information: The power supply voltage is the electromotive force, and the circuit end electric cracking pressure is the electromotive force minus the internal pressure of the power supply, that is, the voltage of the external circuit. The power supply voltage is generally unchanged, and the terminal voltage is determined according to the internal resistance of the power supply.

It is determined by the rolling internal resistance r, the external resistance r and the electromotive force e, that is, i=e (r+r), and then the road-end voltage (Ohm's law of closed circuits) is obtained by the large source width u=ir

Kiss, for you to find these two related formulas.

There are still problems. How do you ask for this?

Okay? Hello, the answer to the second question is 6, and the answer to the third question is 16.

Can you send me the process?

Related information: The formula for calculating the resistance: (1) r = l s (where the resistivity of the resistor is determined by the fissure, l is the length of the resistance, and s is the cross-sectional area of the resistance) (2) Definition:

r=u i(3) total resistance in a series circuit: r=r1+r2+r3+......Total resistance in a parallel circuit of rn(4): 1 R=1 R1+1 R2+......1 rn(5) Sikairan finds resistance by electrical power:

r=u²/p；r=p/i²

Hello, you can look up the formula.

Hello, can you put in the formula to figure it out? Is there still a need for a process now?

Hello, I just accidentally typed a typo. The second question is answered with 5 and the third question is 15.

Summary. The formula for energy is e=1 2cv 2, where e denotes the stored energy, c denotes the capacitance of the capacitor, and v denotes the voltage of the capacitor.

How to find the pro voltage to be described?

Dear<> I can't read it here** You can send it with the content of the text, and the teacher will give you the answer.

Can I send a text The teacher can't receive it here** Forgive me.

Find the voltage. 9V—50 ohms-V1—100 ohms—0V

Pro-Hu is <>

Hello, according to Ohm's law, the current through two resistors is the same when they are connected in series, i = v1 r1 = 9 - v1) r2, where v1 is the voltage value of the beam, and r1 and r2 are the resistance values of the two resistors respectively. Substitute r1 = 50 euros and r2 = 100 euros to get i = v1 50 = 9 - v1) 100

Simplification yields 100v1 = 450 - 5v1

Simplified to get v1 = so the voltage of v1 is volts.

Dear, look at what the teacher analyzed for you, can you understand?

Calculate the reactance of a capacitor of 2 F at a frequency of 50 Hz.

Hello The reactance of a capacitor of 2 F at a frequency of 50 Hz is approx.

How do you ask for it?

The Xinzen reactance formula of the capacitance is: x c = 1 2 fc), where f is the frequency, c is the capacitance value, and is pi. Substituting a capacitance slip value of 2 F and a frequency of 50 Hz into the formula:

x c = 1 2 50 2 10 -6) Therefore, the reactance of a capacitor of 2 F at a frequency of 50 Hz is about .

This one is a 2 f and 50 hz process.

Calculate the energy stored in a 10 uF capacitor when charged to 100V.

Good. The formula for energy is e=1 2cv 2, where e denotes the stored energy, c denotes the capacitance of the capacitor, and v denotes the voltage of the capacitor.

The data is substituted into the formula: e=1 21010 -6100 2= jTherefore, the energy stored in a is joules.

I can read it.

Question 1: How to calculate the actual voltage and actual power The actual voltage and actual power are measured, not calculated. Based on the measured voltage, if the resistance of the load resistance is known, the power can be calculated using p=(u2)r.

However, for some electrical appliances, such as televisions, refrigerators, computers, etc., it is impossible to reach the size of the equivalent resistance, so it can only be measured. If you don't have a power meter, you can measure the voltage and current, and then p=v*i to calculate the power.

Question 2: Calculation and formula of actual power and actual voltage I believe you have mastered the formula. Whether it's finding the actual power or the rated power.

As mentioned earlier, the formula is the same. The key is: whatever you want, you can use whatever conditions you want.

For example, how much is the actual power of a "220V, 100W" lamp connected to a 110V circuit. Speaking of power at 110 volts.

That's why you have to use 110 volts. Then there is p real = u real i real hail = u 2 r. Actually, I don't know if I use p=u real 2.

R can be found by p amount = u amount i = u amount r. You see, the formula is the same, except that when you find the resistance, you have to apply the known conditions p and u to find it. When finding the actual power, use p real and u real to find it.

Try to do it.

Question 3: According to the rated voltage, rated power and actual voltage, how to find the actual power p=w t=ui=(i*i)r=(u*u) r

First, use the rated u and the rated p to calculate the electrical resistance of the electrical appliance: p=(u*u) r

Question 4: How to find the actual voltage p=w t=ui=(i*i)r=(u*u) r

First, calculate the resistance of the electrical appliance with the rated u and the rated p: p=(u*u) r

Question 5: Now that the actual electrical power has been found and the resistance is known, how to find the actual voltage p=u2 r

Hello, the actual voltage = the total current of the circuit * the electrical resistance. When he is in the circuit, because there is resistance, there is a voltage value at both ends of the electrical appliance, and this voltage value is the actual voltage of the electrical appliance in the circuit. The actual voltage is related to the voltage and total resistance of the circuit, and when the actual voltage is greater than the rated voltage, the electrical appliance Qi Zhen is overloaded and may burn out; When the actual voltage is less than the rated voltage, the electrical appliance cannot work normally.

Gao Youcoarse.

Actual voltage Total current of the circuit * electrical resistance. When he is in the circuit, because there is resistance, there is a voltage value at both ends of the electrical appliance, and this voltage value is the actual voltage of the electrical appliance in the circuit.

The actual voltage is related to the voltage and total resistance of the circuit, and when the actual voltage is greater than the rated voltage, the electrical limb is overloaded and may burn out; When the actual voltage is less than the rated voltage, the electrical appliance cannot work normally.

<> in the tributary in the figure above, it is clear that the voltage U=5V and the fiber vertical current I=1A are non-correlated positive directions.

u=uab=5v。

According to kvl: UAB = UAM + UMN + UNB = 5V.

uam=u=？

umn=-3×1=-3（v）。The voltage umn and the current i=1a are non-correlated positive directions, so: umn=-i r.

unb=-4v。

Therefore: u+(huicode-3)+(4)=5, which u=12(v) in front of the vertical.

First, it can be regarded as the following closed loop. Reed.

Second, the current in the loop is constant, which is 1A of the output of the current with the source of Zen knowledge. Raids.

Third, use the loop voltage law (KVL):

5+4+3×1-u=0

u=12(v)

These components are connected in series in the same circuit, so the current is the same source. According to Kirchhoff's voltage law, at any given moment, the sum of the electromotive forces along any circuit in the circuit is equal to the sum of the voltage drops across the resistors. This makes it possible to solve for the pulse output voltage.

u=5+4+3=12v

The formula of the node voltage method is: effective self-conduction x current node voltage effective mutual conductance x adjacent electric voltage = current algebra and specific method flowing into the node, first take the 0 potential point, that is, the lowest node. Take u1 as an example.

Effective self-conductivity is the effective mutual conductance (i.e., inverse proportional) of all the resistors connected to that node, that is, because the conductance is in phase with the other nodes. A portion of the resistor connected to the lead wire flows into the nodal current, i.e. writes directly to the voltage source or current source connected to the circuit. The voltage source is divided by the resistance on the wire.

I will write this letter for you, and you will find the distortions, and the answer to the information on the Internet is the same. It seems that my efforts have been in vain and I can't solve the system of equations.

i=p/u.It can be calculated that the current through the small bulb is i=6 6=1a, and the circuit is connected in series, so the current on the resistor is also 1a, and the voltage is 24-6=18v, and the resistance value of the resistor can be calculated as r=u i=18 1=18 ohms, and the power of the resistance is 18w, which is too big, you only have many resistors and resistors to use, it is recommended to use 24v bulbs directly, if you don't use 4 6v bulbs to string them together, you can also do it, but the power becomes 4 times that of a single one.

Method 1: p u i

p 6W (output power of small bulbs).

U 24v is i total

r1 rx (r1 is the impedance of your current bulb, rx is the resistance that needs to be connected in series) r1 u2 p

i.e. r1 36 6 6

U Total i Total R Total.

i.e. 24 rx = 24 4 6 102

That is, if 102 is connected in series, the output of the small bulb is 6W, but it is recommended that the series connection is larger than this resistor, so as to ensure that the bulb will not burn out.

Method 2: Directly divide the voltage of 24-6=18V to be safe!

Small bulb p=U2R

The resulting resistance is r=u2 p=24*24 6=96 ohms.

From the series partial voltage, ur=6V ur=18v ur, r=ur r, the resistance value of the series should be r=ur ur* r=3*96=288 ohms.

The conditions for this question are insufficient.

In series voltage division, the voltage ratio is equal to the resistance ratio. r1：r2＝u1：

U2 now u1 = 6v, u2 u-u1 16 6 10v so. r2 (u2 u1)r1=(5 3)r1 must calculate the bulb resistance r1. If the lamp power is p, then r1 u 2 p

6V6W bulb with a rated current of i=p u=1a

The voltage of the resistor is U=24-6=18V

So the connected resistor r=u i=18 ohms.

Voltage is the reason why the free charges move in a direction to form an electric current in a circuit. Definition of potential difference (voltage difference): the charge q moves from point A to point B in the electric field, and the ratio of the work done by the electric field force Wab to the charge quantity q is called the potential difference between the two points of AB (the difference of the electric potential between the two points of AB, also known as the potential difference), which is expressed by UAB, then there is a formula:

Where, WAB is the work done by the electric field force and Q is the amount of electric charge. At the same time, it can also be defined in terms of electric potential: If the magnitude and direction of the voltage do not change with time, it is called a steady voltage or constant voltage, which is referred to as the DC voltage, which is represented by the capital letter u.

If the magnitude and direction of the voltage change over time, it is called a fluctuating voltage. For circuit analysis, one of the most important fluctuating voltages is the sinusoidal alternating current voltage (abbreviated as alternating voltage), which changes periodically in magnitude and direction according to the sinusoidal law over time. The instantaneous value of the AC voltage is indicated by the lowercase letters u or u(t).

The device that provides voltage in a circuit is a power supply. Unit: The principal unit of voltage in the SI system of units is volts (v), abbreviated as volts, denoted by the symbol v.

1] 1 volt is equal to 1 joule of work done for every 1 coulomb charge, i.e. 1 v = 1 j c. Strong voltages are often measured in kilovolts (kv), and weak voltages can be measured in millivolts (mv) and microvolts (v).

The calculation process is given after acceptance.