When the switches are all closed, R 1 is short-circuited, and R2 is connected in parallel with the bulb.
The voltage at both ends of the bulb is the power supply voltage. Because the bulb emits light normally, the bulb voltage is
, is 6V.
So the power supply voltage is 6V.
The light bulb current is 0.6A (it can be read or calculated, so it emits light normally).
R2 current I2 =1.8-0.6 =1.2a.
The voltage of R2 is equal to the power supply voltage of 6V.
R2 resistance R2 = 6/ 1.2 = 5ω.
In order to minimize the current, it is necessary to increase the circuit resistance, because R2 and the bulb are connected in parallel, so R2 and the bulb have only one access circuit, which is connected in series with R 1. The bulb resistance is greater than R2 resistance, so the bulb is selected to be connected to the circuit.
because
Is 0-3V, and the voltage of R 1 is 3V, so the bulb voltage is 3V.
At this time, the bulb current is 0.5A (see bulb.
At this time, the resistance of the bulb is 6 ω),
The resistance value is also 6 ohm,
The minimum current is 0.5A
In the circuit shown in the figure, the power supply voltage is constant, the word "10WW" is marked on the bulb L, and r = 10Ohm. When R= 10 is turned off and the slider P of the sliding rheostat slides to the A end, the light bulb L can just emit light normally. When S 1 is turned off and the slider p slides to the b end, the current is expressed as 0.2 A, and the equation is:
(1) According to the mathematical formula of P=, the resistance of the bulb is:
RL= mathematical formula = mathematical formula = 20ω;
(2) When S 1 is turned off and the slider P of the sliding rheostat slides to the A end, the circuit is a simple circuit of the bulb L,
The light bulb emits light normally,
Power supply ∴ voltage U=UL= 10V,
When the S 1 disconnect slider P slides to the B end, the bulb is connected in series with the maximum resistance of the sliding rheostat, and the ammeter measures the current in the circuit.
According to ohm's law, the total resistance in the circuit is:
R total = mathematical formula = mathematical formula = 50ω,
The total resistance in the series circuit is equal to the sum of the individual resistances,
∴ Maximum resistance of rheostat:
R ' = R total-RL = 50ω-20ω= 30ω；
(3) When S 1 is closed and the slider P slides to the A end, the bulb is connected in parallel with the resistor R,
∫ The voltages at both ends of each branch in the parallel circuit are equal,
∴ electric power consumed by resistor r:
PR= mathematical formula = mathematical formula = 10w.
Answer: (1) The filament resistance of bulb L is 20 Ω;
(2) The maximum resistance of rheostat R' is 30 Ω;
(3) When S 1 is closed and the slider P slides to the A end, the electric power consumed by the resistor R is 10W. ..