Xiaohong uses the resistance value to design an electric power point heater with three levels: mediu

When R1 works alone, the electric power can be obtained according to the formula P=U2 R (U2 refers to the square of U, I will not hit the square of U), P1=220*220 121=400WThe working current is i=p u, that is, i=400 220, which is about the same reason, it can be calculated that r2 works alone when the electric power is 800w, and the working current is.

When the switch S1 is closed at low level, the electrical power of S2 is 400W, when the trunk current is medium, the electrical power of the switch S2 is 800W, and when the switch S1 and S2 are closed at the same time when the dry current is high, the electric power of switch S1 and S2 is 1200W, and the trunk current is.

Working in low gear, the switch of the R1 circuit is closed, and the switch of R2 is not in harmony. Working on the high grade, both S1 and S2 are closed. Use your phone for your question, the words are too small to read clearly about the calculation, when the low gear:

Current, voltage, and resistance. Power, voltage, current. High-grade, that is, the method of calculating the parallel resistance.

1) When S1 is closed in low gear, the current I = 220V R1 and the electric power P = 220V square R1

2) When the high grade is high, S1 and S2 are closed at the same time. Electrical resistance r=r1*r2 (r1+r2) current i=220v r.power p=220v.square r.com.

Junior High School Physics Electricity Problems I. Ohm's Law and Its Deformation Formula.

i＝u/r u＝ir r＝u/i

2. The calculation formula of electrical work (or electric energy consumption).

w＝uit＝u2t/r＝i2rt＝pt

w 50r 2500r kw·h or 500imp 5000imp kw·h, w w at the end of the month w at the end of the previous month, etc.

3. The formula for calculating electrical power.

p＝w/t＝ui＝u2/r＝i2 r

4. The formula for calculating the heat generated by the electric current.

q＝i2rt＝uit＝u2t/r＝pt＝w

5. The relationship between the seven physical quantities of the series circuit (taking two pure resistors R1 and R2 in series as an example).

1 Equiquantitative relationship.

i＝i1＝i2 u＝u1＋u2 r＝r1＋r2

t＝t1＝t2 w＝w1＋w2 p＝p1＋p2 q＝q1＋q2

2 Assignment Relationships.

i1 i2 1 1 u1 u2 w1 w2 p1 p2 q1 q2 r1 r2 (partial pressure principle).

6. The relationship between the seven physical quantities of parallel circuits (taking two pure resistors R1 and R2 in parallel as an example).

1 Equiquantitative relationship.

u＝u1＝u2 i＝i1＋i2 1/r＝1/r1＋1/r2

t＝t1＝t2 w＝w1＋w2 p＝p1＋p2 q＝q1＋q2

2 Assignment Relationships.

u1 u2 1 1 i1 i2 w1 w2 p1 p2 q1 q2 r2 r1 (shunt principle).

R1 works alone, and the electrical power is p1=220*220 121=400W. according to the formula p=uThe operating current is i=p u, i.e., i=400 220 approx.

In the same way, it can be calculated that the electric power of R2 is 800W when working alone, the electric power of closing switch S1 and disconnecting S2 when the working current is low-grade is 400W, the electrical power of closing switch S2 and disconnecting S1 when the dry current is medium-range, and the electric power of closing switch S1 and S2 at the same time when the dry current is high-level, and the electrical power of closing switch S1 and S2 at the same time is 1200W, and the dry circuit current is.

First, calculate the power p=66000 300=220wp=u 2 r=u 2 220=220w, and the solution is u=220v

When the current passing through it is, p=i 2r=

Power consumption within 3min: w=pt=55*3*60=9900j