The two lights of A and B are marked with the words 12V 36W and 24V 18W respectively, which are conn

The brightness of a light bulb is determined by its actual power.

Answer: It is known that U A = 12v P A = 36W u B = 24v P B = 18W u Real = 36V

Seeking: p A real p B real.

Solution: A: p=ui iA=pA=pA=36W12V=3ar=U iR=uA=12V3a=4 euros.

The resistance value is unchanged iA4 = U r=36v 4 ohms = 9ap A-real = U A-real * i A-solid = 36v*9a=324w B: p=ui i B = P B = 18w 24v = r = u i r = u B = 24v ohm.

The resistance value is unchanged iB solid = u B real r = 36v 32 ohm = p B real = u B real * i B real = 36v *

p A solid p B solid.

The first light is relatively bright.

Solution: According to the meaning of the topic, it can be known

r A = the square of u A P A = 12 * 12 36 = 4 euros.

r B = u B squared p B = 24 * 24 18 = 32 euros.

Because the current is the same, r A is less than r B.

Therefore, r B gets more voltage than r A, so B is brighter.

rA = 4r B = 32

Because i is the same.

r B is greater than r A.

Therefore, B points to the voltage more.

So B Liang.

The brightness of a bulb is determined by its actual power.

From the known, from r = u 2 p r a = 4 ohms, r b = 32 ohms, discussed with p = i 2r, the two lamps in series current are equal, and r a is less than r b, then, p a is less than p b.

Therefore, the B lamp is brighter.

Of course, it is also possible to use p=ui to discuss, r A is less than r B, then u A is less than u B, the current is equal, p A is less than p B, and it can also be judged that B is brighter.

It is important to choose the appropriate formula for judging such a problem).

First of all, you need to understand that the brightness of the bulb is determined by the actual power. p=, by comparing the magnitude of the two resistors, it is easy to get that the resistance of A is 4 ohms, and the resistance of B is 32 ohms.

In the series circuit, the current through the electrical appliances is equal, with p = the square of the current multiplied by the resistance, it is easy to find that the actual power of lamp B is greater than lamp A, because the actual power of lamp B is greater than lamp A, so lamp B is brighter.

I don't know if you understand that?

From r = u 2 p r a = 4 ohms, r b = 32 euros, so u a = 4 v, u b = 32 v

From p=ui to get p A, so B lights up.

The series current is the same, according to the formula p=ui, the voltage is larger, and it is B bright.

First find the resistance value of the two lamps, and then connect the respective voltages and the actual power obtained,,1The resistance of a lamp, p=u*u r so 15=36*36 r A r A = ohms, 2The resistance of the B lamp, r B = ohms.

3.The two lamps are connected in series in a circuit with a voltage of 36V, the actual voltage and current of the two lamps can be found, you can see that the resistance value of lamp B is greater than that of lamp A, so the voltage on lamp B is higher than the voltage on lamp A, because it is connected in series, so the current of the two lamps is the same, according to p= ui, the actual power of lamp B is greater than that of lamp A, and the rated power of lamp B is also smaller than that of lamp A, so the answer should be d,

d, known by p=u 2 r, the resistance of lamp A is small, and the resistance of lamp B is large. Since it is a series p real = i 2r knows, the actual power of lamp B is large, and the faster the power is consumed, so the brighter lamp B is.

The resistance of lamp A, p=u2 r, so 15=362 rA rA=ohm, the same reason: the resistance of lamp B, r B = ohmic series, the current of the two lamps is equal by p=i2r, so the actual power of lamp B is large, so lamp B is brighter.

The voltage divided in series is also large if the resistance is large.

Because they are connected in parallel in a 6V circuit, their voltages are all 6V, so the rated voltage power of the first lamp is 3W, because p=ui, so the current is resistance=u, i=12

The rated current of lamp B = p u = 1a resistance = u i = 12 because lamp b does not reach the rated voltage, so the actual current = 6v 12 = actual power = u i = 3w

Because time can be canceled, the power ratio is equal to the actual power ratio, the power ratio = 1:1, the resistance ratio = 1:1, the current ratio = 1:1, the actual power ratio = 1:1

r=v^2/p

ra=12 ohms.

RB = 12 ohms.

ra/rb=1

Parallel ua=ub

So the current ia=ib

Actual power PA=PB

The pure resistance of the electric leakage of the lamp is 12 ohms.

B lamp electric pants which resistance is 6 euros.

Then the actual current is 12 back to the stove (12 + 6) = 2 3A The rated current of the lamp is .

Lamp B is 1A, so the current is small when connected in series.

Because less than 2 3a

So it's not normal.

Current = Voltage Resistance.

Power = Voltage Current = Voltage Voltage Resistance.

Resistance = Voltage Voltage Power.

Resistance of lamp A: 12 12 9 = 16 (Jane Gao Ou Qingru) Resistance of lamp B: 24 24 36 = 16 (ohm).

Because the resistors are the same, they are connected in series and connected to a 36V circuit, and the voltage is 36 2 = 18 (volts) for each lamp

The result is e: the nail lamp burns out.

Solution: The resistance of the two lamps is:

R1=U1 P1=12 9=16 (Europe) R2=U2 P2=24 36=16 (Europe) After connecting in series, the actual voltage of the two lamps of A and B is as follows:

u1'=r1u (r1+r2)=16 36 (16+16)=18 (volts).

u2'=u-u1'=36-18=18 (volts) Then, the actual power of the two lamps is :

p1'=u1'r1 = 18 16 = watts) p2'=u1'r1 = 18 16 = watts).

The 12V 9W resistor is 12 12 9 = 16

The 24V 36W resistance is 24 24 36 = 16

After series connection, each gets 18V, so the fiber finch lamp is burned out, the circuit should be broken, and the hole macro has no merit to consume the vertical book rate.

1. The resistance of lamp A r1 = uxu r=12x12 9 = 16 ohms, the resistance of the family hail shirt lamp B r2 = 24x24 36 = 16 ohms, r1 = r2

2. Connect them into the 36V mountain circuit after connecting them in series: because their resistance values are equal, their bearing pressure is also equal, that is, each lamp is bearing pressure 36 2 = 18V, then (result: the first lamp is burned).

The resistance of lamp A is r1=u p1=1210, and the resistance of lamp B is r2=u p2=807

After series connection, the power of the first lamp at this time p1 = i * r1, p2 = i * r2, because r1 > r2, so p1 > p2 that is, the first light is brighter.

The total power of the circuit is P=220 (1210+807)=24W, and then the power after parallel connection of 220V voltage is 40+60=100W

Why do students come here every day to find out?

As can be seen from the question, the resistance of the two lamps is 24 24 = 24 and 20 16 = 25 respectively

It can be seen that the one marked with 20V16W is brighter, this is because the two lights are connected in series in the 24V circuit, the current is certain, the voltage is larger than the resistance, and the power is large, so the 20V16W is brighter than the 24V24W.

Under normal circumstances, the B light is brighter.

Because the impedance of lamp B is higher than that of lamp A, the voltage obtained in series connection will be relatively higher.

Select the resistance of AA: RA = U A P A=3 3=3 A's rated current (current upper limit): IA = U A RA = 3 3 = 1A B's resistance:

r B = u B P B = 6 12 = 3 B rated current (current upper limit): i B = u B r B = 6 3 = 2a after series connection, circuit current: i = u (r A + r B) = 9 (3 + 3) = visible, i i A, i i B.

Therefore, lamp A burns out, and lamp B glows but is darker than normal.

rA = U2 p = 9 3 = 3 ohms;

r B = U2 P = 36 12 = 3 ohms.

Therefore, if the two series connections and the partial voltage are equal, then the rated voltage higher than 3V will be burned out, and A is selected