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The resistance of a 25 watt bulb on a lighting circuit is 1936r, 1, find the voltage U? 2. If a 60 watt bulb is replaced on this circuit, what are the resistance and current of the bulb?
If 25 = u ^ 2 / 1936u = 220V is replaced by 60W bulb, r = 220 ^ 2 / 60 = 807 Ω, I = 60 / 220 = 0.27A
When the electric iron is connected to the lighting circuit, the resistance is 1936 ohm, so how long can it work continuously and consume one kilowatt
The power consumption should be one kilowatt hour. Kilowatt hour is the energy unit
The effective voltage of lighting circuit is 220 v
Therefore, the power of electric iron is p = 220 * 220 / 1936 = 25W
That is, the power of electric iron is 25 watts
So a kilowatt hour can last 40 hours
Connect the bulb marked "2.5V 0.3A" to the 6V power supply, and connect an ohm resistance to the bulb to make it shine normally?
Since the rated voltage of the lamp is less than the supply voltage, it is obvious that a resistor should be connected in series, and the resistor in series should be set as R
The resistor R must share the voltage U1 = U-U lamp = 6-2.5 = 3.5 v
So r = U1 / I lamp = 3.5 / 0.3 = 11.7 Ω
The resistance of small bulb L1 is 18 Ω, and the allowable current is 0.3A. The resistance of small bulb L2 is 12 Ω, and the allowable current is 0.4A,
(1) If the lights are connected in series, what is the maximum allowable voltage? What is the voltage at both ends of L2? (2) if the two lights are connected in parallel, what is the maximum allowable voltage? What is the current through L1?
(1) If the lights are connected in series, the maximum allowable voltage is u = I1 (R1 + R2) = 0.3A * (18 Ω + 12 Ω) = 9V
At this time, the voltage at both ends of lamp L2 is U2 = i1r2 = 0.3A * 12 Ω = 3.6V
(2) U1 = i1r1 = 0.3A * 18 Ω = 5.4v, U2 = i2r2 = 0.4A * 12 Ω = 4.8V. If two lamps are used in parallel, the maximum allowable voltage is 4.8V
The current through equal L1 is I1 ′ = 4.8v/18 Ω = 0.27A
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