A junior high school electrical problem, sincerely seeking analysis!

1. It becomes larger, because when the height of the person to be measured increases, the resistance of the sliding rheostat increases, so the voltage on both sides increases, and the voltmeter is connected in parallel with it, so the voltage is measured, so the number of voltage representations increases.

2 Because the component should work safely, and the current through R2 can not be exceeded, so R total = U total I total =, because R1 = 5, so R2 is at least connected to 9 -5 = 4And because the maximum range of the voltmeter is 3V, the maximum voltage at both ends of R2 is 3V, so the voltage at both ends of R1 is, so I1=U1 R1=, so R2=U2 I2=3V, so the R2 resistance range is 4 -10

When the height increases, the resistance value of R2 becomes larger, the total resistance of the series circuit becomes larger, and the circuit current becomes smaller, so that the voltage of R1 becomes smaller, and the sum of the voltages on the two resistors of the series circuit does not change, so the voltage of R2 becomes larger, that is, the number of voltage representations becomes larger.

Range solution: Since the maximum passing current of R2 is, the minimum total resistance of the circuit is ohms, so the minimum of R2 is 9-5=4 ohms.

Because the maximum voltage of R2 measured by the voltmeter is 3V, the minimum voltage of R1 is and the minimum current of the circuit is 3 ohms.

Let the power supply voltage be U, the bulb resistance r1, and the line electrical slow balancing cover which beam r2 because the ratio of the bulb power before and after = 25:16

So the ratio of the current passing through the bulb = 5:4

Because the power supply voltage does not change, R1:(R1+R2)=4:5, so R1:R2=4:1

The ratio of the power of the series circuit = the ratio of the resistance to the material, so the line resistance is 4W

It must be d, 25-16 is how much!

Answer: (1) According to the question, it can be seen that the current reading is inversely proportional to the percentage error, so? So? =50%

2) The smaller. 3) One-half.

The three points of DCE are located at half of the voltage drop between AB, that is, the three points of DCE are equivalent, and there is no voltage between DC and CE, that is, there is no current, so these two lines can be removed when calculating the resistance, and what remains is a simple series and parallel connection. The final result is 1 (1 2r + 1 2r + 1 2r) = (2 3)r

i1=9i2---i1（ra+r1+r2）=i2(rb+r1+r2)--36/(ra+r1)=81/(rb+r1)--

R1:R2=1:6 is obtained

2. From (6+I1R2) (9+I2R2)=1 to get I2R2=6 so U=9+I2R2=15V

If you use a 12V battery with a 24V10W bulb, then what is its current power?

The resistance of the filament is obtained by P=UI and I=U R, and R=U P is obtained

rl=uo po=(24v) 10w=, get p=u r from p=ui and i=u r to find the actual power of the bulb:

p=u²/rl=(12v)²/

Solution: Let the total voltage be u; When S1 is closed and S2 is disconnected, the current is I1, the voltage at both ends of R0 is U01, and the voltage at both ends of R1 is U1; When S1 is disconnected and S2 is closed, the current is I2, the voltage at both ends of R0 is U02, and the voltage at both ends of R2 is U2.

1) According to p=u*i and the meaning of the title, p1=u1*i1=p2=u2*i2=2w (1 formula).

Since i1:i2=2:1, substituting 1 into it, we get u1:u2=1:2, i.e., 2*u1=u2 (2).

According to p=u r, u1 r1=u2 r2, substituting formula 2 into it gives r1=r2 4 (3 formula).

i1=u (r0+r1),i2=u (r0+r2), that is, [u (r0+r1)] u (r0+r2) ]2:1 Substituting 3 into it simplifies :

r0 ：r2=1：2

2) According to the above title, according to p=u r and u0:u2=r0:r2, p2=4u 9r2 = 2w, that is, u r2 = 9 2w (4 formula).

According to the title, when the sliding vane is located at the midpoint of R0, the total resistance is R2+R0 2= , P=U, and the 4 formula is substituted into it, and the total power P=18 5 W can be obtained.

When set at point A, the sliding rheostat voltage U(A), R1 voltage U(R1A), R2 voltage U(R2A).

When set at point B, the sliding rheostat voltage U(B), R1 voltage U(R1B), R2 voltage U(R2B).

The series circuit has r1 r2=u(r1a) u(r2a)=u(r1b) u(r2b)=k (introduced unknowns).

List 4 equations u(a)+u(r1a)=11 . Equation a

u（r1a）+u（r2a）=6。。。Equation B

u(b)+u（r1b)=10。。。Equation c

u（r1b)+u（r2b）=9。。。Equation D

Subtract the voltage of R2 with k and push the equation b to U(R1A)*(1+K)=6

Equation d pushes out u(r1b)*(1+k)=9

u(r1a) u(r1b=2 3....Equation e

The equation a+b pushes out u(a)+(2+k)u(r1a)=17... Equation f

The equation c +d pushes out u(b)+(2+k)u(r1b)=19 . Equation g

The power supply voltage remains unchanged, u(a)+u(r1a)*(1+k)=e=u(b)+u(r1b)*(1+k).

Equation g-f=u(r1b)-u(r1a)=2 .Equation h

Equation h and equation e solve the system of equations, u(r1b) = 6 u(r1a) = 4

Substituting Equation b u(r2a)=2 Substituting Equation a u(a)=7

So the supply voltage = 4 + 2 + 7 = 13V

Isn't it a list of equations and then mathematically eliminated elements?..

AB point is the sliding rheostat at both ends?

1 Solution When the normal luminescence, the voltage at both ends is u=220v, the current is i=p u=40 220=, and the resistance is r=u, squared, p=(220*220), 40=1210 ohms.

Abnormal light emitting under 110V voltage, and the current flowing I=U2 R=110 1210=

Actual power p=u2 square r=10w

2 Solution Use r=u squared p to find the resistance of two bulbs r1=3 ohms and r2=ohms.

So the total current is I=U (R1+R2)=6 (3+) when connected in series on a 6V power supply

The actual power of bulb 1 p1 = i square * r1 = 16 3w the actual power of bulb 2 p2 = i square * r2 = 8 3

If it is connected in parallel, the rated voltage is reached on the 3V power supply, so it emits light normally.

3 Solution The rated current of the bulb is i=p u=3 6=

To work properly, the resistors in series should share the voltage of 8-6=2V.

So the resistor to be connected in series r=u1 i=2 ohms.

4 Solution: Use i=p u to calculate the rated current of the two bulbs are i1=2a, i2=3a because it is connected in series, so in order to make the lamp not damage, the maximum allowable current in the circuit is 2a, and the allowable voltage u=i1*(r1+r2)=2*(64 16 +144 36)=16v

5 Solution Because the power is the same, it is the same bright when working normally, and R1 = 12 ohms and R2 = 3 ohms are calculated by using r=u square p, and the current is equal when connected in the same circuit, and the power p with large resistance is obtained according to p = i square * r, so r1 is bright.

The electrical energy consumed by the bulb is converted into electrical energy and light energy, and the electric energy consumed by the electric soldering iron is all converted into heat energy, because they consume the same electric power, so the electric soldering iron produces more heat at the same time.

When the filament is attached, the length becomes shorter, the resistance becomes smaller, and the power p becomes larger according to p=u squared r, so it is brighter.

1。When"220V 40W" bulb glows normally, then the voltage at both ends of the bulb is 220V, the current is 40 220, and the electric power is 40W

If it is connected to 110V, does the bulb glow normally and how much current flows? What is the actual power?

It can't emit light normally, and the bulb resistance = 220 * 220 40 = 1210 ohms.

So at 110V, the current = 110 1210 = 1 11A

Power = 110 * 1 11 = 10W

2。"3V 3W" and 3V 6W"Two bulbs are connected in series on a 6V power supply, find the total current, the actual power and voltage of the two bulbs? If connected in parallel on a 3V power supply, will the two bulbs glow normally?

Resistance of 3W = 3*3 3 = 3 ohms.

6W Resistance = 3 * 3 6 = Ohms.

So the total current = 6 (3+.)

3W voltage = 3*4 3=4V, another = 6-4=2V

Normal glow in parallel.

3。A "6V 3W" bulb is connected to an 8V power supply, in order to ensure normal operation, how much resistance should be connected in series or parallel?

Bulb current = 3 6 =

Resistance divider = 8-6 = 2V

Resistance = 2 ohms.

4。The existing lamp L1 "8V 16W" and lamp L2 "12V 36W" are connected in series in the circuit, in order to keep the lamp from damage, what is the maximum allowable current in the circuit? What is the maximum allowable voltage?

L1 rated current = 16 8 = 2A

L2 rated current = 36 12 = 3A

So the maximum current is 2A

At this time, the normal working voltage of L1 is 8V, and the L2 voltage = 12*12 36*2=8V, so the maximum voltage is 16V

5。"6v 3w"and "3v 3w.""Which one is bright to work properly? Which one is lit in series on the same circuit?

220V 660W "Electric Heating Iron and"220V 660W" electric lamp, which produces more heat?

The electric light at home is broken, the filament has to be connected, and the electric light is brighter than before, why?

The resistance of 6V is larger, so the partial voltage is larger, so the power is greater, so it is brighter.

After connecting the filament, the filament becomes shorter, and the resistance will become smaller, so the power will become larger, so it will be brighter.