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The power supply voltage is 4V, and the resistance value of resistance R is 10 ohm. After the switch S is closed, the resistance R will release 10 joules in 100 seconds The bulb is connected in series with a resistor The power supply voltage is 4V, The resistance R is 10 ohm, After the switch S is closed, 10 joules of heat is released from the resistor R in 100 seconds It is known that the rated power of lamp L is 1.2W. (lamp resistance remains unchanged) Find the rated voltage of L?
The resistance power is P1 = 10 / 100 = 0.1W
Current I = under root sign (P1 / R1) = 0.1A
Partial voltage on resistance U1 = IR1 = 0.1 * 10 = 1V
The partial voltage of electric lamp is U2 = 3V
The bulb resistance is 30 ohm (the ratio of bulb to resistance R is equal to the ratio of partial voltage on the two devices)
The rated voltage of bulb is u = root (P2 * R2) = root (1.2 * 30) = 6V
As shown in the figure, the resistor R with a resistance value of 10 Ω is connected in series with the electric lamp L with a rated voltage of 3.8V, and then connected to both ends of the 4V power supply. Close the switch s, and the electric power of the electric lamp L is & nbsp; 0.4W______ Ω, the voltage at both ends is______ 5. The current in the circuit is______ A. The rated power of lamp L is______ W. (regardless of the effect of temperature on resistance)
According to P = UI: the current through the bulb I = plul, because the bulb and resistance are in series, the voltage at both ends of the resistance ur = u-ul, then the current through the resistance I = URR = u − ULR, so plul = u − ULR, finishing 0.4wul = 4V − ul10 Ω, so ul = 2V, circuit current I = plul = 0.4w2v = 0.2A, resistance r two
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