The lamp and a resistor R are connected in series and then connected in a circuit with voltage of U. the current flowing through the lamp is I, and the electric power consumed by the lamp is?

The lamp and a resistor R are connected in series and then connected in a circuit with voltage of U. the current flowing through the lamp is I, and the electric power consumed by the lamp is?

P total = P Lamp + P resistor
P total = u · I, P resistance = I ^ 2 · R
P lamp = u · I - I ^ 2 · R

The power supply voltage is 3V, and the resistor R = 20 Ω is connected in series with the lamp L "6V, 3.6W". After closing the switch, (1) the current in the series circuit
The power supply voltage is 3V, and the resistor R = 20 Ω is connected in series with the lamp L "6V, 3.6W". After closing the switch, (1) what is the current in the series circuit? What is the voltage at both ends of R? (2) what is the actual power consumed by the lamp L?

1. The resistance of the lamp is: R lamp = u ^ 2 / P = 6 × 6 / 3.6 = 10 Ω
Series current: I = u / R total = 3V / (20 Ω + 10 Ω) = 0.1A
The voltage of R is: U & # 39; = IR = 0.1A × 20 Ω = 2V
2. The voltage of lamp is: u lamp = 3V - 2V = 1V
The actual power is: P '= u lamp I = 1V × 0.1A = 0.1W

When the bulb connected to the home circuit is normally emitting, the current passing through the bulb is 0.2A. What is the resistance of the bulb at this time?

According to Ohm's law, r = u / I = 220 V / 0.2A = 1100 Ω

In the series circuit, the power supply voltage is 6V, the switch is closed, the current expression number is 0.3A (1) the resistance value of the small bulb (2) the electric power of the small bulb

R=U/I
=6/0.3
=20 euro
P=UI
=6*0.3
=1.8W