Find the problem of Ohm s law and electric power of medium difficulty

Definition formula for electrical power:

p=w/t。

Derivation formula: w=iut, p=w t=uit t=ui.

And according to Doomming Ohm's law: i=u r

There are: u=ir, p=ui=iti=i and p=ui=u u r=u r.

Answer: n-electric power u-voltage r-resistance i-current.

n=u squared r n=i squared r

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.

Electrical power1 Definition: The work done by an electric current per unit of time.

2 Physical Meaning: Representation.

Electric current does work. Fast and slow physical quantities. The brightness of the bulb depends on the actual power of the bulb.

3 Electrical power calculation formula: p=ui=w t (applicable to all circuits) for pure resistive circuits.

It can be derived: p=i2r=u2 r

Series circuits. The commonly used formula is: p=i2r

p1：p2：p3：…pn=r1：r2：r3：…: rn parallel circuit.

The commonly used formula is: p=u2 r

p1：p2=r2：r1

Regardless of the electrical appliances.

Series or parallel. Calculate the total power.

The common formula is p=p1+p2+....pn

4 Units: SI units.

Watts (w) commonly used units.

Kilowatts (kW).

Let the bulb resistance value be rl, yes.

The bulb is connected in series with R1 rl*(V (R1+rl)) 2 =3 (1).

The bulb is connected in series with R2 rl*(v (r2+rl)) 2 =3 4 (2).

From 3w 6v 6v rl, find rl 12 known condition r2=n r1

1), (2) divide the two equations, and (n-2)r1-12=0 is solved to obtain n with two values that meet the constraints, i.e., 4 and 3

n 4, r1=6, r2=24, the voltage source should be 9v;

n 3, r1=12, r2=36, the voltage source should be 12v

The conditions are not complete, right?? For example, is the power supply voltage constant?

Amount 2 p=12,rl*u (r1+rl)=6v,rl*u (r2+rl)=3v,(r2+rl)=2, (r2+rl)=2, since r1:r2=1:n, let r1=x, i.e., nx+12=2*(x+12),(n-2)*x=12,n is a positive integer less than 5 12=1*12=2*6=3*4, so n=3,4

x=12,r1=12,r2=36

n=4,x=6,r1=6,r2=24