A few physical questions about voltage resistance

1。If there is only a voltmeter and no other components, it is not possible to talk about series or parallel, but there is no doubt that the voltmeter can directly measure the power supply voltage if the range allows.

2.In the circuit, the voltmeter can be regarded as a segment-open switch, because its resistance is very large, and the bulb naturally does not light up.

3.The voltage is provided by the power supply, and it is impossible for the switches and appliances themselves to have voltage. It's like you can't prevent water from coming out of a faucet without a water source.

4.Same as the principle of the 3 questions.

5.And the power section is on, there is no voltage, the same principle as the 3 questions.

6.The position of the components in a series circuit can be arbitrary, and the action is independent of the order, you are right.

I teach physics, and QQ is: 464398817

Frequently answer students' questions through QQ.

1. There is only a voltmeter in series in a circuit, will this burn out the power supply? What about parallels??

No, not in parallel.

2. In a circuit, if only the voltmeter and the bulb are connected in series, will the light be on?

It depends on what the power supply voltage is!! If you're in junior high school, treat the light bulb as if it's not on.

3. Do the switches and bulbs themselves have voltage?

There is no pull. 4. If there is no current in a circuit, then the bulb has no voltage?

Yes, if there is no voltage (and to be channeled) there is no current.

Switch --- bulb - (positive) battery (negative) - disconnected--- above is a simple circuit diagram (the drawing is ugly, it will be a moment, the four corners are right angles) In this diagram, can the bulb measure the voltage? ?

If you can't understand your picture, add me to draw a picture.

6. In a series circuit, there are sliding rheostats, batteries, switches, and bulbs. Is it possible to adjust the brightness of the bulb no matter where the rheostat is placed?

Yes. But connect it properly.

Otherwise, if you add mine, I will also be the first 2.

1. Neither case, the voltmeter is equivalent to an approximate infinity resistor.

2. No, the reason is the same as above.

3. I don't know what you mean, the voltage is provided by the power supply, if it is on the circuit that is turned on, there is a partial voltage on the bulb, but there is no voltage itself.

4. Yes. 5. No, open circuit.

6. Pure series circuit, yes, as long as the switch is closed.

1.The series connection will not burn out the power supply, because the voltmeter itself is high resistance! Without thinking! Think of the voltmeter as a broken switch! Parallel connection is even more impossible!

2.No! The reason depends on 1 problem!

3.Not! If you don't believe it, you can use a voltmeter to measure!

4.No! There are counterexamples, but I forgot, I'm sorry!

5.I can't understand the picture.

6.Yes. No matter the location!

1. The series will not be paralleled.

2. If the voltmeter resistance is very large, the light is not on.

3. The resistance of the switch is very small, and it is negligible when calculating. The resistance of the bulb is sometimes counted.

4. If there is no current flowing, there is no voltage formation.

5. How to measure the scale without a watch hand?

6. When adjusting the sliding rheostat, you should also consider whether the bulb will be burned out.

a+bi(1)。

The traditional method of analyzing an active linear two-terminal network using Thevenin's theorem is to first find the open-circuit voltage followed by the equivalent resistance, and finally draw the Thevening equivalent circuit to find the quantity to be found. This process is cumbersome because it is a step-by-step solution and a step-by-step circuit diagram is drawn, especially for two-terminal networks with controlled sources, which is more complicated and error-prone.

The imposition excitation method is now commonly used to find the equivalent resistance of a passive linear two-terminal network with a controlled source. Finding the linear equation of the port's response and excitation allows you to quickly find the two parameters of the Thevening (or Norton) equivalent circuit, which is much easier than the traditional step-by-step solution.

Extended Information: Notes:

1. The reference direction of the UOC is consistent with the reference direction of the open-circuit voltage of the two-terminal network.

2. Find the open-circuit voltage UAB of the two-terminal network, then UOC=UAB

3. Remove the source of the above-mentioned active two-terminal network, and obtain the equivalent resistance rab of the passive two-terminal network, ro=rab.

4. Use UOC and RO in series to form an equivalent voltage source, connect it at both ends of the branch to be sought, form a single-loop simple circuit, and find the current or voltage in it, which is the request. Let the right end of the resistor with a resistance value of 2 be point C. According to the partial circuit Ohm's law of stool tremor = u defeat r can be known:

i=i1+is=3+2=5（a）。

The principal unit of voltage in the SI unit system is volts (v), abbreviated as volts, which is represented by the symbol v. 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 conversion relationship between them is:

1kv=1000v。

1v=1000mv。

1mv=1000μv。

Because the currents are equal, the circuit is connected in series. To L2 to emit light normally, so the voltage is 6V, according to the elegant proportion, the voltage of L1 is 3V, and finally find the power supply voltage, the voltage in the series circuit is U1 + U2 = U total, so the power supply voltage is 9V.

Bulb L2 is.

6v3w, i.e. l2.

When 6V emits light normally, the ratio of voltage is 1:2, and the two bulbs are connected in series, and the voltage is divided in series, 6 2 = 3, that is, the voltage on the bulb L1 is 3V, so 6 + 3 = 9V

When the supply voltage is 9 volts, L2 emits light normally.

The current is 1 to 1

It can be seen that they are connected in series.

The rated current of L2 is:

Therefore, the current in the circuit is.

The L1L2 resistance is 6

12 voltages are.

Because L2 emits light normally, the voltage of L2 is 6V.

Because the ratio of voltages is 1:2

Therefore, the voltage of L1 = L2 voltage * 1 2 = 6V * 1 2 = 3V because the bulb is connected in series in a certain circuit.

So the current is equal, so the supply voltage = L1 voltage + L2 voltage = 3V + 6V = 9V So when the supply voltage is 9 volts, L2 emits light normally.

Resistance of L1 lamps.

R1 = uu1 p1 = (6v 6v) 6w = 6 resistance of l2 lamp.

R2 = UU Amount P = (6V 6V) 3W = 12 The rated current of the L1 lamp.

i1 = p1 u1 = 6w 6v = 1a

The rated current of the L2 lamp.

I = P = 3w 6v =

The two lamps are connected in series, and the current is equal, which is equal to the small rated current.

Supply voltage u=i(r1+r2)=

The L2 lamp emits light normally, and the current passing through it is equal to its rated current.

A section of resistance wire, when the voltage at both ends is 10V, the current is (50 ohms) when the voltage is reduced to 5V. When the voltage is zero, the resistance is (50 ohms).

Resistance is an inherent property of circuit components, which is only related to its own material, structure, environment and other factors, and will not change with the change of voltage and current in the circuit.

According to the voltage and current to find the resistance is 10, because the resistance will not change, so it is still 50 ohms, and both blanks are filled in 50

Solution: r=u i=10

Since the resistance is independent of the magnitude of voltage and current.

Hence added. 5v、0v

The voltage is also 50

r=v/i

So when the voltage is 10V and the current, the resistance r=10 ohms can be calculated.

When the voltage is reduced to 5V, the resistance is still 50 ohms, and if you ask about the current, you can calculate it according to the formula, the current is.

When the voltage is zero, the resistance is still 50 ohms. The current is also O.

A conclusion is attached to you. From the above answers, it can be seen that the current is directly proportional to the voltage, the higher the voltage, the greater the current, and the smaller the voltage, the smaller the current.

The reason is simple.

As can be seen in the question, these three voltage time periods are 10V, 5V, and 0V, and the same resistor is used.

Resistance is the physical quantity in a substance that hinders the flow of electric charge, i.e., the value of resistance.

For the resistor in the ideal state, the resistance value of the resistance does not change with whether there is a voltage added to the outside world, but the required resistance value must be obtained according to the conditions given in the question, such as obtained by the deformation of u=i*r, or r= *l s( : resistivity, l: resistance length, s:

resistance cross-sectional area).

So the answer to this question is 50

1. The smaller the resistance of the high-voltage transmission line, the smaller the electrical energy lost, and the resistance of the metal wire is inversely proportional to the cross-sectional area of the conductor and proportional to the length of the conductor. Therefore, when the metal wire is thick and straight, the resistance is small and the power loss is small.

2, q i rt, the greater the resistance, the more heat generated, the same as above, the thinner and longer the wire, the greater the resistance. Because the thick and straight wire has low resistance and generates less heat, it is not used.

The size of the resistance is related to the thickness, the coarser the resistance, the smaller the resistance, and vice versa, the greater the forehead resistance, the more heat it generates, so you should understand why! Hope, handwritten o( o

Because the thinner the wire, the greater the resistance, r=

l s, p is the resistivity, l is the length, s is the section area of the wire, from which it can be seen that the thicker the transmission line, the better, so that the line loss is less, that is, the power consumed on the transmission line is less, p=i 2r, in order to save electric energy!

The two forms of the AB formula, P=UU R, are suitable for a certain voltage, such as comparing the current and power force rate of several silver towns or parallel resistors, which is simple; P=IIR, which is suitable for the case of several resistors in series (where the current is the same), compare the voltage and power of the resistors, and simply travel to the bridging.