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The small bulb L marked with "6V, 3W" and the resistor R with the rated power of 6W are shown in the figure. When the switch S is closed, the lamp l will light normally, and the current indication is 0.75a 1. The value of resistor R 2. The actual power consumed by the whole circuit 3. Xiaoqiang's solution is as follows According to Ohm's law, the resistance of resistor R = u / I = 6V / 0.75a = 8ohm 2. The actual power consumed by the resistor R is PR = 6W, and the actual power consumed by the whole circuit is p = PR + pl (small L) = 6W + 3W = 9W, 3. Please point out the mistakes in Xiaoqiang's solution,
When the switch S is closed, the lamp L lights normally, that is, the small bulb gets 6V voltage, consumes 3W of power, and the current flowing through it is I = P / u, that is, 3 / 6 = 0.5A, then the current flowing through the resistance is 0.75-0.5 = 0.25A, and the power generated by the resistance is 0.25 * 6 = 1.5W (P = IU), the resistance is U / I = 6 / 0.25 = 24 ohm
The power consumption of the whole circuit is 1.5 + 3 = 4.5 watts
A resistor R with a rated power of 6W is not a known condition, but its power should not exceed 6W, otherwise it will burn out the resistor
Connect the 6v3w small bulb l with the 6W resistor R as shown in the figure, close the switch s, the lamp L lights normally, and the current indication is 0.75a
1 find the resistance value of resistance R
The actual power consumed by the whole circuit
Can not see the diagram, the circuit should be parallel
Rated current of 6v3w small bulb:
I＝P/U＝3/6＝0.5(A)
Current consumed by resistor R:
I＝0.75－0.5＝0.25(A)
1》 Calculate the resistance value of resistance R:
R＝U/I＝6/0.25＝24(Ω)
2》 Actual power consumed by the whole circuit:
P＝UI＝6×0.75＝4.5(W)
In the series circuit, the resistance R = 20 ohm, the lamp L is marked with "6V, 3.6W", after closing the switch s, the bulb will light normally;
(2) What is the current in the circuit? (3) what is the voltage at both ends of resistance R? (4) what is the power consumed by resistance R?
1、 6*6/3.6=10Ω
2、 3.6W/6V=0.6A
3、 20Ω*0.6A=12V
4、 12V*0.6A=7.2W
Connect R1 and R2 in series and then connect them to a 12V power supply. If R1 = 20 ohm, the voltage at both ends of R2 is 9V
What is the resistance of R2?
The current in series circuit is equal, i.e. I = U1 / R1 = U2 / R2
U1 is the pressure drop on R1, U2 is the pressure drop on R2, and we can know from the relationship in the question that U2 = 9V, U1 = 12-u2 = 3;
Therefore, there are:
3/20=9/R2,
R2 = 20 * 9 / 3 = 60 Ω
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