A simple circuit problem, a circuit problem.

The electrical work p is the square of i*r

One is 8 and the other is 2, and 8 is 4 times that of 2, so if the power is equal, the square of i corresponding to the large resistance should be 4 times the square of i corresponding to the small resistance, so the i corresponding to the large resistance should be 2 times the i corresponding to the small resistance. Parallel connection is a shunt, one is 8, one is 2, if the current is 2 times the relationship, the resistance should also be, so connect a 2, the total is 4, on the line. Note that the power of the original two resistors is equal, not that the total power of the two resistors is equal to the power of the other resistor after series connection.

Substitution method, count once:

a：r1=8ω，r2。=4ω

Then the voltage of R1 = 2 times the voltage of R2 (he is divided into the general voltage to give another resistance) current i1 = u R1 = U 8

i2=u/(r2+2)=u/4

So i1=i2 2

p1=u*i1

p2=(u/2)*i2

p1=p2

p=i^2r=u^2/r

p1=u^2/8

p2=u^2/2

A resistor with a resistance value of 2 is connected in series with R2.

i=u/r=u/4

p=i 2r=(u 2 4) 2*2=u 2 2=p12Please pay attention to the formula.

If you want to use the multiple-choice question, use the elimination method: if you add 6, then the sum of the power of r2 and rx is equal to r1.

If you want to calculate, it will be as follows:

By equal power, there is i2:i1 2

rx+r2＝r

r1i1＝ri2

1. Normal luminescence, all are 40W, so 1:1

2. i=p u so 1:2

Since it is normal light, it can be considered that the two lamps A and B are connected to the circuit of no more than 110V. Now let's say it's a 110V circuit.

A = 40 * 110 2 220 2 = 40 4 = 10w, p A p B = 1:4

A i B = p A p B = 10 40 = 1:4

A r B = p B P A = 40 10 = 4:1

Option A solution: 1) When the switch S is closed, the switch is disconnected, and the slider P slides from the B end to the midpoint, the indication of the ammeter changes, and the indication of the voltmeter is 6V; Because the voltmeter measures the voltage of the bulb at this time, which happens to be the rated voltage, the current flowing through the bulb and the sliding rheostat is i=p u=3w 6v=

ib= (ib is the current flowing through the original sliding rheostat) and rl=u2 p=(6v)2 3w=12 (rl is the resistance of the bulb) (rl+r)ib=(rl+r 2)i

12ω+r）× r/2）×

r=8 so the supply voltage u=ib rtotal=

2) When the position of the sliding blade p is kept unchanged, the switch is closed, and the indication of the ammeter changes by 2a, then i total =, then r and r 0 are connected in parallel, so the current flowing through r is ir = u (r 2) = 8v 4 = 2a so the current flowing through r 0 is i 0 =

So r0=u i0=8v

SolutionWhen S is closed and S1 and S2 are disconnected, the circuit is a series circuit composed of L and R, and the voltmeter measures the voltage at both ends of L, and the ammeter measures the current in the circuit.

It is known that the bulb L is marked with "6V 3W", which gives R=U2 P=36 3=12 ohms.

Because the current in the series circuit is equal everywhere, the current in the circuit is i=u r=6 12=

Power supply voltage u = lamp +. 5r---1

When the slide p slides from the B terminal to the midpoint, the indication of the ammeter changes.

Power supply voltage u = lamp +. 4r---2

It can be concluded that the supply voltage is 8V, r = 8 ohms.

Keep the position of the slider p unchanged, close S1, S2, the circuit is a parallel circuit at this time, and the lamp L is short-circuited.

In the circuit, R0 is connected in parallel with R, then the voltmeter does not have an indicator, and the ammeter measures the total current.

Keeping the position of the slider P unchanged and closing S1 and S2, the indication of the ammeter changes by 2A

One of the resistance values becomes smaller, the total resistance becomes smaller, so the current must become larger, so it should be added 2a, that is, the total current of the parallel circuit is, note that at this time, p is at the midpoint, and the resistance value is r 2

Total current i=u r0+u (r 2)=

U=8V, R=8 ohms, substituting can find R0=16 ohms.

There are a few words in your question that I don't quite understand: "What does the degree of voltage mean by not knowing?"

I'm guessing your original question was "The voltmeter reads as." Here's what to ask now.

Step 1: Take the reading of the ammeter:

Since it is a three-phase symmetrical circuit, now let the RMS value of one of the phase currents be i(a), which is the reading of the ammeter in the figure.

The next three loads z are regarded as an independent star-connected three-phase circuit, the line voltage of this part is known, and the phase voltage is equal to one-third of the root number of the line voltage, so the phase voltage = can be obtained

And the modulo of z is equal to (15 2 + 3 * 15 2) and then open square = 30

So the phase current (also the line current i) is equal to: 660 30 = 22A. This is what the ammeter reads.

The second step is to find the line voltage between AB.

1. Find the total impedance of zl and z.

Total resistance = 16 ohms;

Total inductive impediment = (2 + 15 3 )j (i.e. 2 + 15 root number 3) j

So the total impedance (modulus) = (16 * 16) + (2 * 2 + 60 3 + 15 * 15 * 3) re-square = (256 + re-square =

2. Find the voltage drop u of the current i on the upper impedance

u=22a*

3. The above is the phase voltage. And the line voltage is equal to 3 times the root number of the phase voltage:

So the line voltage UAB = end.

Resistance 6V 12 when emitting light normally

The current passing through l is so the current passing through r is.

So the power of R2 is UI=

Disconnect S2L in series with R1.

The resistance of l = U square p=36 3=12

Current i=6 (12+12)=

Power p= i, square r=

1.The supply voltage is 6V and the lamp resistance is 12 ohms.

for 20 euros. 3.The actual power of the lamp me.

Listen well in class and read well. These are the topics in the books.

Solution: 1. Only the bulb is connected to the circuit, the bulb L emits light normally, and the power supply voltage is the resistance of the bulb R1=U 2 P=(6V) 2 3W=12 Euro2, at this time, the circuit is L and R2 in parallel, and the current through the lamp L i1=U R1=6V 12 ohms = through R2 current i2=i-i1=

R2 actual power p2=ui2=6v*

3. At this time, the lamp is connected in series with the rheostat R3.

The total resistance in the circuit r=r1+r3=12 ohms+12 ohms=24 ohms, the actual current i=u r=6v, 24 ohms=

Lamp l actual power p=i 2r1=(ohm=

This group of people who answered c are all ruining people......

At the moment of shut-off, the current source output is 2A, and the capacitance IC = -4A, and I = 6A.

The correct one is b, because the output current of the current source is constant, i(0+) = 2 + 4 = 6a.