Why is the resistance and power of electric furnace high in parallel circuit?

Why is the resistance and power of electric furnace high in parallel circuit?

The wrong argument is that the power is small when the resistance is large. When the voltage of the parallel circuit is constant, the smaller the resistance is, the greater the branch current is, and the more work the current does to the consumer

The relationship between AC power and DC power

It is suggested that R1 is connected to the power supply in parallel with the resistance of the sliding rheostat
According to the characteristics of parallel circuit, the difference between ammeter A2 and A1 is the current intensity passing through R2. R1 and R2 can be calculated by Ohm's law, and the ratio of heat released by R1 and R2 at the same time can be calculated by electrothermal law
101. (L) 15 ohm. 40 v
(2) 6 a. 32 v. 3073 J
It is suggested to analyze the circuit first: the sliding rheostat is connected to the part R'2 of the circuit in series with R 1, and then connected to the power supply, then the reading of the ammeter is the current flowing through R 1 and R'2, and the reading of the voltmeter is the voltage applied to both ends of the resistance R'2, or the voltage value separated from both ends of the power supply by R'2 connected in series with R 1
On the basis of the above analysis, R'2 can be obtained by applying Ohm's law. According to the voltage distribution characteristics of series circuit, the power supply voltage should be equal to the sum of the voltages on R1 and R'2
When the slide plate slides to point a, R'2 = R 2, that is, the resistance of the slide rheostat connected to the circuit is the largest
Ohm's law is used to calculate the current passing through R1 and R2 in this case, and the reading U2 of the voltmeter is also calculated. Then the heat generated in one minute is calculated according to Joule's law
102. (1) 500 watts. (2) 96.8 ohm. (3) 413.2 watts
It is suggested that the actual voltage of an electrical appliance in normal operation should be equal to the rated voltage, so the actual voltage of the electric water heater in this question should be 220 v. after 12 minutes of operation, the indication of the watt hour meter will increase by 0.1 degree, that is, 0.1 degree of electricity consumption, It is worth noting that the actual voltage is no longer the rated voltage, so the actual power at this time is definitely not the power calculated in question (1), but the resistance of the water boiler can be considered unchanged,
103. (L) 0.5 A. (2) 120 ohm
The key to the problem is to make clear the circuit connection when the key is opened and closed
(1) When K1 and K are disconnected, R2, bulb L and ammeter are connected in series, as shown in Figure 73
(2) When K1 and K2 are closed, bulb L is short circuited, so R1 and R2 form a parallel relationship, and the ammeter is connected in series in the trunk road, as shown in Figure 74
In addition, we also need to pay attention to the analysis mentioned in topic (1): when K1 and K2 are disconnected, l just lights up normally. The meaning of l just lights up normally is: at this time, the actual voltage applied at both ends of L is the rated voltage value, the actual power of L is equal to the rated power, and the actual current passing through L is the rated voltage value
(2) When K1 and K2 are closed, R1 and R2 form a parallel circuit
The supply voltage u can be obtained from the above solution
104.4 ohm. 16 ohm
When the rheostat contact P slides to the leftmost end, the voltage at both ends of the bulb is the highest and can not exceed 6 V, so the voltage at both ends of the bulb must not be small. When the sliding head P slides to the rightmost end, the voltage at both ends of the bulb is the lowest, but can not be less than 3 V. according to
4 degrees
It is suggested that this is not only a problem of calculating electric power, but also a very practical application in daily life
According to the title, there are four 60 watt fluorescent lamps in each classroom, and 24 × 4 60 watt electric lamps in 24 classrooms, so the total power of the building can be calculated. Then the power saving degree in a month can be calculated by w = P · T. attention should be paid to the use of units in calculation, P should use kilowatts, t should use hours
The total power of electric lamp in teaching building is p = 24 × 4 × 60W = 5.76kw
One month power saving time t = 0.5 × 30 = 15:00
One month power saving w = Pt = 5.76 kW × 15 h = 86.4 kWh (KWH)
106. (1) 403.3 ohm. (2) 0.76 degree

About electric power calculation
How long does it take to heat 2kg water from 20 ℃ to 100 ℃ with 2kW electric furnace? The efficiency of electric furnace is known to be 60%, and the specific heat capacity of water is 4.2 * 10 ^ 3j / (kJ. K)

Q=ηPt
t=Q/(ηP)=cm(t2-t1)/(ηP)=(4.2*10^3*2*(100-20))/(0.6*2000)=560s

Calculation of electric power
A 6v6w light bulb and a 6v3w light bulb are connected in series in a 6V power supply. What is their respective actual power? If they work for 1min, what is the total electric energy consumed by the circuit? What is the heat? If they are connected in parallel with a 6V power supply, what is the actual power, Which light bulb is brighter? If parallel connected in 3V power supply, calculate their actual power? And please bring the formula! You can't only know the answer! For example, the formula of u = IR, q = uit and so on!

(1) Let 6v6w be L1, 6v3w be L2. P1 = u & sup2 / r1ri = 6p2 = u & sup2 / r2r2 = 12wr1: R2 = U1: u2u1: U2 = 1:2ui + U2 = 6u1 = 2 U2 = 4p1 (actual) = u & sup2 / R1 = 0.67wp2 (actual) = u & sup2 / R2 = 1.33w (2) q = w = u & sup2 / R total. T = 120J (3) L1 bright (4) p1 = u & sup2 / R