There is a bulb with a resistance of 15 ohm. When the two ends of the bulb are applied with a voltage of 6 V, the bulb can work normally. If the bulb is connected to a 7.5 V power supply, it will still light normally. What is the resistance of the bulb in series?

There is a bulb with a resistance of 15 ohm. When the two ends of the bulb are applied with a voltage of 6 V, the bulb can work normally. If the bulb is connected to a 7.5 V power supply, it will still light normally. What is the resistance of the bulb in series?

The normal working current of the bulb is I = U1 / r = 6 / 15 = 0.4, and the total resistance after series connection is r = U2 / I = 7.5 / 0.4 = 18.75, so the resistance that should be series connected is r = 18.75-15 = 3.75

There is a bulb with a resistance of 15 Ω. When 6V voltage is applied at both ends of the bulb, the bulb can work normally. If the bulb is connected to a 7.5V power supply, how large a resistance should be connected in series to make the bulb shine normally? What is the power of the bulb at this time?

When the bulb lights normally, if the voltage at both ends is 6V, then u ′ = 7.5v-6v = 1.5V; I ′ = I = ur = 6v15 Ω = 0.4A; R ′ = u ′, I ′ = 1.5v0.4a = 3.75 Ω; P = UI = 6V × 0.4A = 2.4W. A: the bulb should be connected with a 3.75 Ω resistor in series; at this time, the power of the bulb is 2.4W

How to connect a 12V 8W bulb and 6 12V 4W lamps in a 24 V power supply

Take two 4-watt in parallel and 8-watt in series; the remaining four 4-watt in series (two circuits), and finally connect the three circuits in parallel to the power supply

In the circuit as shown in the figure, the power supply voltage remains unchanged, the resistance of the constant value resistor is 4 Ω, and the maximum resistance of the sliding rheostat is 10 Ω. When the slide P is placed at the midpoint of the sliding rheostat, the voltage representation is 8V; when the slide moves s from the midpoint to the B end, the voltage representation becomes 6V; when the slide moves s from the midpoint to the a end, the voltage representation is ()
A. 4VB. 6VC. 10VD. 12V

According to Ohm's law, I = ur = 8v4 Ω = 2A; up = IRP = 2A × 5 Ω = 10V; power supply voltage: U0 = u + up = 8V + 10V = 18V. When the slide moves s from the midpoint to the B end, the voltage representation becomes 6vi ′ = u ′ r = 6V4 Ω = 1.5A; up ′ = u0-u = 18v-6v = 12V; RP ′ = up ′ I = 12v1.5a = 8 Ω Distance s, resistance change 3 Ω, when slide moves distance s from midpoint to a end, resistance of access circuit becomes 2 Ω. I ″ = u0r + RP ″ = 18v4 Ω + 2 Ω = 3A; u ″ = I ″, r = 3A × 4 Ω = 12V. So a, B, C are wrong, D is correct, so D is selected