η=W，w total = ghfs = gnf。
When the pulley block is not labor-saving, that is, when F≥G, η = gnf ≤1n;
(2) When the slider P moves to the A end, only R 1 works, and the current is expressed as 0.3A,
Power supply voltage U=0.3R 1
R 1 power consumption is Pa=Ia2R 1=U2R 1? …①
When the slider P moves to the B end, R 1R2 is connected in series, and R2 =15Ω.
According to ohm's law, IB = sum of ur = ur1+R2;
At this time, the power consumed by R 1 is Pb = ib2r1= (ur1+R2) 2r1… ②.
Pa: Pb = 9: 4, namely u2r1:(ur1+R2) 2r1= 9: 4.
The solution is r 1 = 2R2 = 30ω, which can be obtained by substituting r 1 = 30ω into ① ② equation.
U=9V，Ib=0.2A
② When the slider P moves to the B end, the voltmeter measures the voltage of R2;
According to ohm's law, U2 = ir2 = 0.2a× 15ω = 3v.
A: The resistance of (1)R 1 is 30 Ω; (2) When the slider P moves to the B end, the voltmeter indicates 3v.