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R1 = 2 ohm, R2 = 6 ohm. After the switch is closed, the indication of ammeter is 4.8A Solution: the voltage ratio of resistance R1 and R2 Ratio of current through r1.r2 supply voltage The circuit is as follows: starting from the positive pole and passing through the switch, it is divided into two channels, one is R1, ammeter a, return to the negative pole, one is R2, ammeter a, return to the negative pole
1) Ur1: UR2 = 1 parallel circuit voltage equal
2） Ir1：Ir2=(Ur1/Rr1):(Ur2/Rr2)=(Ur1/2):(Ur2/6)=3:1
3) Rtotal = 1 / (1 / rr1 + 1 / r2) = 1 / (1 / 2 + 1 / 6) = 1.5 Ω
U total = I total * r total = 4.8A * 1.5 Ω = 7.2V
Two resistors R1 = 5ohm, R2 = 10ohm in parallel in the circuit, close the switch, the number of ammeter is 1.2V, find the total current in the circuit
The ammeter is connected in series with R1 and the reading is 1.2A
I1=1.2A
U=I1R1=1.2 * 5=6V
I2=U/R2=6/10=0.6A
I = I1 + I2 = 1.2 + 0.6 = 1.8A. Total current
(the voltage of parallel circuit is equal, and the total current is equal to the sum of branch currents)
(the ammeter in series with whom will measure the current)
The current intensity of a lightning stroke is about 20000 a, the voltage is 1 billion V, and the discharge time is 0001 seconds. The average power of a lightning stroke is several thousand watts
How many joules of work does the current do? If 1% of this energy can be used, how long can a 100 watt lamp work?
Good answer, plus points! Fast! It's better to write known, seek and solve
Known: I = 20000A, u = 10 ^ 8V, t = 0.001s
How long does the electric power P and 1% of the time energy supply 100W electric lamp
P=UI=2*10^12W
t=Q*1%/100=2*10^12*0.001*1% / 100=2*10^5s
On average, the current intensity of a lightning is about 2, the voltage is about 1 billion V, and the discharge time is about 0.001 s
How many kilowatts is the average power of a lightning stroke. How much work does the current do
P=UI=1X10^9VX2A=2X10^9W
W=Pt=2x10^9wx0.001s=2x10^6J
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