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If resistance R1 is connected to a circuit, the power consumed is P1; if another resistance R2 is connected to the same circuit in series, the power consumed by R2 is P2. If P1: P2 = 4:1, R1 "select" greater than "," less than ", or" equal to "R2
It's wrong. Sorry, it's changed
Because P1: P2 = 4:1, it can be assumed that R2 = n × R1, so the second current I2 = I1 / (2 ×√ n) can be calculated
Because the voltage of the two circuits is the same, we can get the same result
I1×R1=（N+1）×R1×（I1/（2×√N））
(n + 1) / (2 √ × n) = 1
The solution is n = 1
So R1 = R2
Two resistors R1: R2 = 1:2; when connected in parallel to the power supply, what is the ratio of power consumption P1: P2?
In parallel, their voltage U is equal. P = u ^ 2 / R, so their power P1 / P2 = R2 / R1 = 2
Two resistors R1 and R2 are connected in series under the same supply voltage, and the power consumption is P1 and P2 respectively. Now they are connected in parallel under the same supply voltage
Find the power P2 'consumed by R2 at this time
P2=u2*I2
Given U2, R2, find I2
I2=U2/R2
p2=U2*I2
That's OK
As shown in the figure, in the circuit, the power supply voltage U remains unchanged, if the electric power of resistance R1, R2, R3 and R4 is P1, P2, P3 and P4 respectively
R1 = 8 Ω R2 = 16 Ω R3 = 8 Ω R4 = 16 Ω
The incorrect one in the following relation is ()
A.P1>P3,P1P4
C.P2-P1>P3+P4
D.P2-P1
The answer is D:
By decomposing the graph, we can get that R3 and R4 are connected in parallel first, then Rx is obtained, and the parallel resistance is 5.333 ohm
Then the resistor is connected in series with R1 and R2
In a: because R1 = R3, the current through R1 and R2 is equal to the current through R3 and R4 in parallel, so P1 > P3, the current through R1 and R2 is the same, and R1 is the same
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