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Connect a bulb marked "10V. 5W" in series with a resistor and connect it to a 10V power supply. The power consumed by the resistor is 1W. Calculate the actual power of (1) bulb and (2) bulb
Assuming that the bulb is a pure resistor and the resistance of the bulb does not change with the working voltage, the resistance of the bulb is r = u ^ 2 / P1 = 10 ^ 2 / 5 = 20 ohm
If the loop resistance is 20 + R after series resistance R, the loop current I = u / (20 + R) and the resistance power P2 = I ^ 2 × R, that is:
（U/(20+R））^2×R＝1
Substituting data
(10 / (10 + R)) ^ 2 × r = 1, the quadratic equation R ^ 2-60r + 400 = 0 is obtained
The resistance R is 52.36 ohm,
Loop current I = u / (1200 + R) = 10 / (10 + 52.36) = 0.1382a
be
(1) The voltage of bulb U1 = I * r = 0.1382a × 20 = 2.7369v
(2) The actual power of the bulb is p = U1 × I = 2.7369v × 0.1382a = 0.3820w
The lamp marked with 12V and 6W is connected in series with a constant resistance of 12 Ω and then connected to the power supply. The power consumed by the lamp is 1.5W
(1) What is the voltage at both ends of the lamp? (2) what is the current through the constant resistor? (3) what is the power consumed by the constant resistor?
U=9V I=0.25A W=0.75W
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