About a junior high school electrical question, a junior high school electrical question

Find the lamp resistance r1: according to p=u2 r, r1=(220v)2 100w=484

The actual current of the circuit is measured according to the actual resistance of the bulb: according to P=I*IR, i=9 22A

According to U=IR, the actual voltage of the bulb U1=198V, the wire resistance can be regarded as the resistance R2 in series with the bulb, and the voltage of R2 can be obtained U2=220V-198V=22V, and according to the equal current in the series circuit, the power of the wire is P=UI=22V* 9 22A=9W

p rated = u square r

Find the r-lamp=484

Then use p actual = i square r lamp.

Find the current = 5/11.

Finally, use p total = i u

The total power obtained is 100 watts.

Subtract the power of the lamp and the power of the wire = 100-81 = 19

First calculate the resistance of the bulb, and then calculate the voltage at both ends of the bulb and the current flowing through according to the actual power, the total voltage is subtracted from the voltage at both ends of the bulb to obtain the voltage at both ends of the wire, and this voltage is multiplied by the current to obtain the power consumed on the wire.

You can think of a wire as a resistor, at 200V, the electrical power is 100W, p total = p bulb + p wire, and the power consumed on the wire is 19W

(1) Because R1 and R2 are connected in series, the voltage at both ends of R1 is obtained: U1 = U-U2 = 15V-2V = 13V; The power consumed by R1 is: P1 = U1 I = 13V

2) If 15 is selected to replace R1 access circuit, when the R2 slider slides to the B terminal, the current through the ammeter is: I=U R1=15V 15=1A. will exceed the range of the ammeter.

The ammeter may be damaged. When the slider slides to the R2A terminal, the ammeter reads as follows: I=U (R1+R2)=15V (15 +20 )=, although it does not exceed the range, but the voltmeter reads:

u2=u×r2/(r1+r2)=15v×20ω/35ω=。will exceed the voltmeter 3V range. So, don't choose a 15 resistor.

24 resistors should be selected.

When access 24 replaces R1, make the maximum reading through the ammeter be, and the resistance value of R2 access circuit is obtained: R2=1; When the voltmeter reading is maxed out to 3V, the resistance value of the R2 access circuit:

r2=3v/(12v/24ω)=6ω.

The voltmeter can be connected to another range for the same analysis.

c If the ammeter has no indication, it must have been opened, and the ammeter, L1 has no voltage diagram, and the small bulb L1 is open because it is a series circuit, so according to your words, I can imagine one.

So that the indication of the ammeter and voltmeter can reach the maximum value of a certain range! There's something missing here, isn't there?

Let's discuss it separately and use these two resistors to bring in and calculate.

A No, the switch can be placed on the trunk circuit to control all of it.

b No, the current of the series circuit is equal, and if it is not equal, it is in parallel.

c is in parallel, not in series.

d is the correct answer.

As for the series circuit you mentioned, there is also a situation where the voltage is equal, and it is only when the two resistors are equal (r u i, i equal, u equal).

Normal luminescence means that the actual power of the lamp is equal to the rated power.