The bulb and a resistor are connected in series to the power supply. The power of the lamp is 81w, and the power of the resistor is 9W. If the lamp is directly connected to the power supply, it will light normally Calculate the rated power of the lamp

The bulb and a resistor are connected in series to the power supply. The power of the lamp is 81w, and the power of the resistor is 9W. If the lamp is directly connected to the power supply, it will light normally Calculate the rated power of the lamp

According to P = I2R, we can get P lamp: P resistance = R lamp: R resistance = 9:1. Because u = IR, u lamp: u resistance = R lamp: R resistance = 9:1 in series, u lamp = (9 / 10) U power supply (9 / 10) U power supply voltage. When the bulb is connected to the power supply alone, both ends of the bulb become power supply voltage U

The brightness of the bulb depends on the product of voltage and current at both ends of the bulb------

The product of voltage and current, namely electric power

Does the intensity of current and voltage affect the brightness of bulb?

Of course, it does. The factors that determine the lamp height are:
1. Power P = UI;
2. Brightness of surrounding environment

Figure a is the image of the current in the small bulb changing with the voltage at its two ends. Connect the bulb to the circuit shown in Figure B and close it

First calculate the resistance of the bulb r = u / I = 2 / 0.5 = 4 Ω. From the power of 1W, P = UU / r u = 2V
Then the power supply voltage is 2V, because it is in parallel, the change of ammeter can only become larger, so the current of R is 0.1A
Parallel voltage unchanged P = UI = 2 * 0.1 = 0.2W 1W + 0.2W = 1.2W total power