Ohm s law in physics and electrical power or something, I don t know who can guide the maze

Ohm's law i=u r

Electrical power p=ui=u2r=i2r

There are four physical quantities in total, and what is required must be contained, and that one is constant.

Ohm's Law Current i=u r

Series circuit: i total = i1 = i2 (in a series circuit, the current is equal everywhere) u total = u1 + u2 (in a series circuit, the total voltage is equal to the sum of the voltages at both ends of each part) The corresponding r is calculated according to Ohm's law derived formula r = u i.

Parallel circuit: i total = i1 + i2 (in parallel circuit, the dry current is equal to the sum of the currents of each branch) u total = u1 = u2 (in the parallel circuit, the power supply voltage is equal to the voltage at both ends of each branch) r = r1r2 (r1 + r2).

When there are n constant-value resistors R0 in parallel, the total resistance r=r0 n, that is, the total resistance is less than any one of the resistors, but the more parallels, the smaller the total resistance, the series voltage division (voltage) and parallel shunt (current).

The light bulbs are bright or dim depending on their actual power, not the rated power.

The actual power p=ui is the voltage across the bulb multiplied by the current flowing through the bulb.

When connected in series, the current flowing through the two bulbs is equal, so whoever has the greater voltage at both ends will have more actual power, and know that the voltage U=IR so whoever has the greater resistance will have the greater voltage at both ends.

In summary, the two bulbs are connected in series, and the larger the resistance is brighter.

When connected in parallel, the voltage at both ends of the bulb is equal, so whoever flows through the current is larger, whose actual power is larger, the brighter, and the i=u r voltage is equal, so whoever has a small resistance, the current flowing through whom the current is greater and brighter.

Therefore, when connected in parallel, the smaller resistor is brighter.

Two bulbs in series, let one of them emit light, it must be the one that is allowed to pass through the maximum current is smaller, or not large, the maximum current allowed to pass through is small, it will definitely burn out, so the current is calculated according to the current of the two bulbs that is the smaller of the maximum current allowed to pass.

Then according to the formula u=ir r=r1+r2, you can find the voltage at both ends of the two bulbs.

The maximum is the smallest, and you can look at the minimum value that can ensure the safety of the circuit.

A sliding rheostat directly changes the resistance in the access circuit, thus changing the current.

So, if you ask for the minimum resistance, that is, you can find the maximum current, and you can find the minimum resistance in the circuit based on the maximum current allowed by the bulb or other electrical appliance, and then subtract the resistance of the bulb (or other appliance) to get the minimum resistance of the sliding rheostat.

Ohm's Law. i=u|r

Already. The current in a conductor is directly proportional to the voltage across the conductor and inversely proportional to the resistance of the conductor.

Electrical power formula.

p w tp = ui (defined).

Electrical power is equal to the product of voltage and current (p=u i) 1 watt = 1 joule second = 1 volt ampere Symbolic meaning and unit w - electrical energy - joules (j).

w—kilowatt*hour (kw*h) t—time—seconds (s) t—hour (h).

p - the power of the electrical appliance - watts (w).

p—kilowatts (kw).

Two sets of units, according to different needs, choose the appropriate unit for calculation) The formula for electrical power also has:

p=i2 r p=u2 r (these are the deformation formulas).

Persistence is required, and only those with perseverance will succeed.

Clarify the basic concepts and familiarize yourself with where they come from and how they relate to each other.

It is necessary to do some questions independently, with quality and quantity. There should be a certain number of questions, not too many, not too few, but also a certain quality, that is, a certain degree of difficulty. Anyone who learns mathematics, physics and chemistry will not be able to learn well without going through this level.

Solving problems independently, sometimes slowly, sometimes detours, sometimes even unable to solve them, but these are normal and the only way for any beginner to succeed.

It is necessary to be clear about the physical process, and there is inevitably a hidden danger of solving the problem if the physical process is not clear. Regardless of the difficulty of the topic, you should draw as much as possible, some sketches are sufficient, and some should be precise to show the geometric relationship. Drawing can transform abstract thinking into figurative thinking and grasp physical processes more precisely.

With graphs, state analysis is fixed, dead, and intermittent, while dynamic analysis is alive and continuous.

Study materials should be well preserved, classified, and marked. The study materials are categorized into practice questions, test papers, lab reports, and more. Marking means, for example, for practice questions, that the general questions are not marked, and the good questions, valuable questions, and error-prone questions are marked differently for future reading, and marking can save a lot of time.

It is necessary to attach importance to the knowledge structure and systematically grasp the knowledge structure, so that the scattered knowledge can be systematized. It is as large as the knowledge structure of the whole physics, as small as the knowledge structure of mechanics, and even down to the chapters, such as the knowledge structure of statics, and so on.

1.Keep the formula in mind.

2.Learn to analyze circuits (just do more questions and practice).

3.Do more calculations (there are easy to difficult, but you must learn the textbook first) Ohm's law and electrical power are not so difficult!

I wish you the best of luck soon!!